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Trigonometry

","image":"
\\"\\"

Trigonometry graphic.

"},"right-triangle-trigonometry":{"title":"Right Triangle Trigonometry","sections":[{"text":"

Right Triangle Trigonometry

Alexander DeCarlo

Trigonometry, as in the name, deals with the study of triangles, angles, ratios, trigonometric functions, graphs and a lot more. It is one of the most important topics in Mathematics and is the gateway for numerous other advanced concepts. While math revolves around Trigonometry, Trigonometry revolves around right triangles. A major reason for this is that only right triangles satisfy the Pythagorean Theorem, which states a2 + b2 = c2, where sides a and b are the legs, while side c is the hypotenuse. One of the key aspects of right triangles is their strict vertical and horizontal side due to the 90\xb0 angle. This also means side c is essentially the slope (rise / run) of the triangle. Let\'s take a look at this right triangle:

\'A","image":"
\\"\\"

Diagram of right triangle.

"},{"text":"

With just this diagram, we can identify many different ratios between the sides. There are a total of 6 ratios, each a trigonometric function. These operations are designated as functions because they completely rely on the angle between both sides.

Let\'s focus on angle \u03b1. Starting with the sine function, we know sin = opposite / hypotenuse. Applying that to angle \u03b1, we find that sin(\u03b1) = a / c, because side a is opposite to angle \u03b1, and side c is the hypotenuse of the triangle. Sine works the exact same way with any other angle. For example, sin(\u03b2) = b / c, as side b is opposite to angle \u03b2, and side c is again the hypotenuse of the triangle. Now that we know all about sine, let\'s look at the other core trigonometric functions:

\u03b8 for any given angle:
Sine -> sin(\u03b8) = opposite / hypotenuse
Cosine -> cos(\u03b8) = adjacent / hypotenuse
Tangent -> tan(\u03b8) = opposite / adjacent

Swapping the numerator and denominator of these core functions gives us their reciprocals:



Cosecant -> csc(\u03b8) = hypotenuse / opposite = 1 / sin(\u03b8)
Secant -> sec(\u03b8) = adjacent / hypotenuse = 1 / cos(\u03b8)
Cotangent -> cot(\u03b8) = adjacent / opposite = 1 / tan(\u03b8)

These 6 functions are all extremely important, with relationships to be further elucidated on as we go through trigonometry. For now though, it\'s important to simply understand where these functions come from and how they\'re used in relation to right triangles.

","image":"
\\"\\"

All 6 trigonometric functions of an angle on a triangle of side lengths 5, 4, and 3.

"},{"text":"\'A

Trigonometry will help us solve triangles like these. Our goal is to find side length c, but we aren\'t given both sides a and b. So what is side c? Which trigonometric function will encompass both side length11 and side c to return the angle? Looking at the 42\xb0 angle, we can see the side with length 11 is adjacent to it, and side c is the hypotenuse of the triangle. Going back to the trigonometric functions, cosine conveniently uses both the adjacent and hypotenuse sides of a right triangle.



cos(42\xb0) = 11 / c
c * cos(42\xb0) = 11
c = 11 / cos(42\xb0)

Here cos(42\xb0) is just a constant, as cos(42\xb0) will be the same for all triangles with a 42\xb0 angle. Using a calculator equipped with the cosine function, we arrive at c \u2248 14.8.

","image":"
\\"\\"

Can you find the length of side b? (b \u2248 6.3)

"}]},"right-angle-trigonometry-test-problem":{"title":"Right Angle Trigonometry Test Problem","sections":[{"text":"

Right Angle Trigonometry Test Problem

Alexander DeCarlo\\"A

Solution:Let y = the base of the entire triangle.
tan(30\xb0) = 50 / y
y * tan(30\xb0) = 50
y = 50 / tan(30\xb0)

Let z = the base of the inner right triangle, z = y - x.
tan(60\xb0) = 50 / z
z * tan(60\xb0) = 50
z * tan(60\xb0) = 50

x = y - z
x = 50 / tan(30\xb0) - 50 / tan(60\xb0)
x \u2248 57.7

","image":"
\\"\\"

All 6 trigonometric functions of an angle on a triangle of side lengths 5, 4, and 3.

"}]},"general-angle-trigonometry-i":{"title":"General Angle Trigonometry I","sections":[{"text":"

General Angle Trigonometry I

Alexander DeCarlo

How can you extend the use of trigonometry to non-right angle triangles with trigonometric functions? The key is in breaking up the \u201cGeneral Triangle\u201d into right triangles and seeing what can be done from there. Let\'s look at the general triangle, where every angle, not just two, are unknown.

\\"Standard

By dropping an altitude at the height to the base of the general triangle, we have created two right triangles. Looking more closely, we can observe a relationship between the two triangles.

sin(B) = h / a
sin(A) = h / b

Then, we can solve for the height.

a * sin(B) = h
b * sin(A) = h

Taking advantage of their equality, we can merge the equations into the following:

a * sin(B) = b * sin(A)

And finally, dividing both sides by ab yields us our final result.

sin(B) / b = sin(A) / a

Realizing that we placed the altitude at an arbitrary angle, we could have rotated the triangle and placed it anywhere else for the same relationship. Therefore, sin(C) / c must also be equivalent.

\u2234 sin(A) / a = sin(B) / b = sin(C) / c

This very simple proof establishes the theorem known as The Law of Sines.

So what kinds of triangles can we solve with this?

"},{"text":"

AAS Triangles

Solving triangles refers to finding every angle and side length associated with that triangle. AAS - Angle Angle Side, means you know two angles and a side, with the order of appearance being angle-angle-side. Because the direction of rotation is arbitrary to its properties, SAA is functionally the same thing. ASA means you know two angles and a side in between. Looping back to AAS/SAA triangles, they are the easiest to solve.

\\"Standard

To find missing angle C in the above description, simply subtract the known angles by 180\xb0 as all triangle degrees add to 180\xb0. C = 110. Finding the other sides is simply a matter of using The Law of Sines.

sin(A) / a = sin(B) / b = sin(C) / c, meaning the angle maintains a ratio with its opposite side, enabling us to find a missing angle/side assuming we have its counterpart and a full angle/side pair.

sin(30\xb0) / a = sin(40\xb0) / 11
a * sin (40\xb0) = 11sin(30\xb0)
a = 11sin(30\xb0) / sin(40\xb0), a \u2248 8.56

Solving for angle C would follow the same process.

"},{"text":"

SSA Triangles

SSA Triangles are more tricky to solve, as they lead to ambiguity which could result in two possible ways to create the triangle.

\\"Standard

We can set up:

sin(70\xb0) / 10 = sin(B) / 9
9sin(70\xb0) = 10sin(B)
sin(B) = 9sin(70\xb0) / 10

Here, we must use the inverse sine function. Basically, if sin(\u1340) gives us a side length, then sin-1(side length) gives us our angle. The only thing is, two possible angles satisfy the sine inverse in all cases. When you plug sin-1(9sin(70\xb0) / 10) into the calculator, you will get one value, the PRINCIPLE value. An explanation for all of this will be in later articles regarding trigonometric functions. sin-1(9sin(70\xb0) / 10) \u2248 57.7. Our other possible angle is simply 180 - PRINCIPLE ANGLE, in this case, 180 - 57.7 = 122.3. (For those who understand the unit circle or are revisiting from a later article, we are essentially reaching the equivalent sin(\u1340) value from QII or QIII.)

To see if both of these triangles are possible, let\'s try to get the 3rd value and see if it all makes sense.

180 - 57.7 - 70 = 52.3
180 - 122.3 - 70 = -12.3

We cannot have negative angles, so we can reject the 122.3\xb0 triangle. If it yielded a positive third angle, we would have to solve for the rest of our values for BOTH the triangles, and simply have two possibilities.

"}]},"general-angle-trigonometry-ii":{"title":"General Angle Trigonometry II","sections":[{"text":"

General Angle Trigonometry II

Alexander DeCarlo

The Law of Sines is an important relationship in the general triangle, but there\'s another core relationship we can implement when dealing with triangles, the Pythagorean Theorem. Consider the general triangle with an altitude dropped, creating two right triangles composing the general triangle. From side c, let\'s call x the base length of the smaller triangle and c - x the base length of the larger inner triangle.

\\"Standard

Instead of relying on sine, let\'s use cosine on angle B to try to do something with that x variable.

cos(B) = x / a
a * cos(B) = x

Now we have x in terms of our side lengths and angles of our general triangle. As we can see, we have two right triangles, so we can create some relationships using the Pythagorean Theorem. Note that we want to find an end relationship that doesn\'t contain variables such as h or x, as these aren\'t the side lengths or angles of our triangle.

h2 + x2 = a2
h2 = a2 - x2

h2 + (c - x)2 = b2
h2 = b2 - (c - x)2

To remove h, we can set these two equations equal to one another.

a2 - x2 = b2 - (c2 - 2cx + x2)
a2 - x2 = b2 - c2 + 2cx - x2
a2 = b2 - c2 + 2cx
a2 + c2 = b2 + 2cx

Remember, x = a * cos(B) so to remove x, substitute this in.

a2 + c2 - 2cb * cos(B) = b2

Since this is a general triangle, variable names can be changed and swapped as they are arbitrary, the only thing that has to stay is the relationship between the cosine angle and the side.

Therefore, to make it look better, we can change this to: a2 + b2 - 2ab * cos(\u03b3) = c2. The cosine angle stays opposite to side c, with the previous case opposite to side b.

This is called The Law of Cosines, for its use of the cos function instead of the sin function.

If you notice, The Law of Cosines is actually just an extension of the Pythagorean Theorem! When cos is 90\xb0, cos(90\xb0) = 0. This removes the -2ab term and simply gives us a2 + b2 = c2, which is pretty much a special case of The Law of Cosines.

"},{"text":"

SAS triangles

In this case, The Law of Sines is unusable, leaving us to rely on The Law of Cosines.

\\"Standard

We can find the missing side with The Law of Cosines.

92 + 82 - 2(9)(8)cos(40\xb0) = c2
34.7 = c2
c \u2248 5.9

We know all three sides, so to find an angle we can simply use The Law of Sines. We do NOT have to worry about the possibility of there being two triangles because if you know the lengths of all three sides, there can be only one possibility for what the angles can be.

"},{"text":"

SSS triangles

\\"Standard

To solve SSS triangles, we can use The Law of Cosines. To eliminate ambiguity when solving, solve for the angle of the largest side first. This removes the possibility of a second triangle because solving from the longest side gives us the largest angle.

52 + 62 - 2(5)(6)cos(x) = 72
25 + 36 - 60cos(x) = 49
-60cos(x) = -12
cos-1(12/60) = x
x \u2248 78.5

Proceed with The Law of Sines to get the other angles.

"}]},"the-unit-circle-i":{"title":"The Unit Circle I","sections":[{"text":"

The Unit Circle I

Alexander DeCarlo

To truly understand trigonometry, one must comprehend how angles work, understanding the unit circle, its relation to angles, and a novel angle measurement called radians.

Angles reach a maximum of 360\xb0, completing a full rotation in circular motion. This range from 0 - 360\xb0 exists because 360 is a conveniently divisible number (by 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, etc.). While practical for everyday use, it lacks the mathematical precision we desire.

\\"Circle

Consider a circle with a radius of 1, making the circumference C = 2\u03c0. This equates a full 360-degree rotation to 2\u03c0 radians, a unit used in advanced trigonometry. To convert between radians and degrees, multiply by 180/\u03c0 or \u03c0/180, respectively.

Now, observe the unit circle on the coordinate plane with an angle traversing it:

\\"Circle

Two angles are measured, originating from the right x-axis and extending outward. The blue and red lines, termed terminal rays, gauge the angle size. Notably, the red terminal ray measures externally, while the blue one measures internally, following the counterclockwise motion. For example, angle a is around 135\xb0 or 3\u03c0/4 radians, while angle b is greater. For angles exceeding 360\xb0, they continue counterclockwise, such as 500\xb0 being equivalent to 140\xb0 after a full rotation. Negative angles rotate clockwise; -1\xb0 equals 359\xb0. This insight allows us to divide the unit circle into four quadrants.

\\"Circle

Every \u03c0/4 (90\xb0) defines a new quadrant, with x and y lines acting as quadrantal separators.

Now that we understand this improved angle system and its application around the unit circle, let\'s explore its significance and connection to right triangles. Unlike general right triangles, special triangles like the 45-45-90 and 30-60-90 offer exact angle and side measurements.

\\"Two

These special triangles, created using the Pythagorean theorem and specific base splits, provide exact values without trigonometry. Placing these triangles within the unit circle, with a hypotenuse of length 1, allows us to compress them by dividing by 2 and 3, respectively. This results in three triangles per quadrant: 45-45-90, 30-60-90, and 60-30-90. This process enables the mapping of coordinates for various angles on the unit circle, forming a helpful diagram for memorization.

\\"A

How does this relate to trigonometric functions? The crucial point lies in the fact that the hypotenuse of these triangles is always 1. Thus, sin(\u03b8) = opp, where \\"opp\\" is the y-coordinate, and cos(\u03b8) = x-coordinate. This insight leads to the realization that coordinates on the circle are simply (cos(\u03b8), sin(\u03b8)). For instance, with our special triangle values, sin(\u03c0/4) = sqrt(2)/2.

Examining the unit circle reveals essential insights into trigonometric functions:

  • Sine and cosine values range between -1 and 1.

  • Sin and cosine values are equal at two values: \u03c0/4, -\u03c0/4

  • In QI and QII, sin is positive; in QIV and QI, cos is positive.

  • As sin increases, cos decreases, and vice versa. Two different angles return the same value, either in the top or bottom half for sin and left or right for cos (sin(\u03c0 - \u1340) = sin\u1340, cos(2\u03c0 - \u1340) = cos\u1340).

  • Finally, recognizing that cos(\u1340) = x and sin(\u1340) = y, tan(\u1340), the slope, is found as y/x, making tan(\u1340) = sin(\u1340)/cos(\u1340). Discovering these relationships is termed finding identities, a concept explored further in our trigonometry studies.

"}]},"the-unit-circle-ii":{"title":"The Unit Circle II","sections":[{"text":"

The Unit Circle II

Alexander DeCarlo

Let\'s explore some intriguing relationships between trigonometric functions using the unit circle while maintaining clarity and coherence. First, let\'s examine the coordinates of points on circles with radii greater or less than 1.

sin(x) = y * r
r * sin(x) = y
r * cos(x) = x

Simply multiply by r, the radius. The ratios are still the same.

(r * sin(x), r * cos(x))

Consider the unit circle.

\\"A

Here, four angles are presented with colored terminal rays. The red and blue rays are reflected across the y-axis, and the blue and yellow across the x-axis. If we examine cos(blue) and cos(yellow), they are the same. Yellow is simply the negative angle of blue; thus, cos(-x) = cos(x). With sine, sin(-x) = -sin(x). Examining the relationship between blue and red, or yellow and purple, we observe that sin(\u03c0 - x) = sin(x), and cos(\u03c0 - x) = -cos(x).

We also observe that cos(2\u03c0 - x) = cos(x), and sin(2\u03c0 - x) = -sin(x). Knowing that tan(x) is the slope, when cos or sin is negative, tan will be negative. Therefore, tan(x + \u03c0) = tan(x), tan(x - \u03c0) = -tan(x), and tan(-x) = -tan(x). These relationships are intuitive and can be quickly deduced from the unit circle.

Now, let\'s explore the relationship between sine and cosine.

\\"A

When rotating 90\xb0, cos(x), which is x, would be rotated completely vertically upward, becoming sine, and sine would become cosine. Starting from the triangle in the first quadrant, upon rotation, the cosine and sine values swap places, and the sine becomes negative. This process is repeated in the second triangle, yielding the following relationships:

sin(x + \u03c0/2) = cos(x)
cos(x + \u03c0/2) = -sin(x)
sin(x - \u03c0/2) = -cos(x)
cos(x - \u03c0/2) = sin(x)

If we examine the tangent values, the cosine and sines are swapped, and one value becomes negative. Adding a right angle, \u03c0/2, yields the negative reciprocal of the original. Therefore:

tan(x + \u03c0/2) = tan(x - \u03c0/2)

More succinctly:

tan(x + \u03c0/2) = -cot(x)

These relationships showcase the beauty of trigonometric functions.

"}]},"the-unit-circle-iii":{"title":"The Unit Circle III","sections":[{"text":"

The Unit Circle III

Alexander DeCarlo

While we are acquainted with the representations of cos and sin on the unit circle, it\'s essential to explore how all trigonometric functions manifest and derive identities from these visual connections.

\\"A

Acknowledging that cos and sin correspond to our x and y values, the question arises: how do we express tan as a length measure? Leveraging the unit circle\'s triangle with a hypotenuse of 1, we can establish a right angle from the (cos, sin) point.

\\"A

Upon closer inspection, we observe that tan represents that line, given that tan is opp/adj, and adj is 1. This elucidates the name \\"tan,\\" signifying its tangent relationship to the circle. Additionally, it visually explains why tan(\u03c0/2) and tan(\u03c0/4) are undefined, extending infinitely without intersecting the x-axis. To determine sec, representing adj/opp, we leverage the triangle formed with tan, realizing that sec must denote the hypotenuse of that triangle.

\\"A

Applying a similar process, we can ascertain cotangent.

\\"A

Similarly, the derivation of cosecant follows a comparable procedure.

Exploring the relationships between trigonometric functions through the Pythagorean theorem, we unveil crucial connections:

sin2(x) + cos2(x) = 1
tan2(x) + 1 = sec2(x)
cot2(x) + 1 = csc2(x)
sin2(x) + (sec(x) - cos(x))2 = tan2(x)
sec2(x) + csc2(x) = (tan(x) + cot(x))2

These relationships are commonly known as the Pythagorean Identities. Among them, the most significant is sin2(x) + cos2(x) = 1, a fundamental identity that merits in-depth study.

"}]},"using-the-pythagorean-identity-sine-and-cosine":{"title":"Using the Pythagorean Identity: Sine and Cosine","sections":[{"text":"

Using the Pythagorean Identity: Sine and Cosine

Alexander DeCarlo

Let\'s review the previously derived Pythagorean Identity: sin2(x) + cos2(x) = 1.

Given: sin(\u03b8) = 2/3, \u03b8 in QII

\\"A

22 + x2 = 32
4 + x2 = 9
x2 = 5
x = \xb1sqrt(5)

cos(\u03b8) = adj/hyp
cos(\u03b8) = \xb1sqrt(5) / 3

Cosine must be negative in QII, therefore we reject positive sqrt(5) / 3.

x = -sqrt(5)

While this represents one method of solving, leveraging the Pythagorean Identity provides a more streamlined approach.

sin2(\u03b8) + cos2(\u03b8) = 1
(2/3)2 + cos2(\u03b8) = 1
cos2(\u03b8) = 5/9
cos \u03b8 = \xb1sqrt(5) / 3

Reject positive sqrt(5) / 3.

cos(\u03b8) = -sqrt(5) / 3

Then we can get the adjacent side from how we calculate cosine:

cos(\u03b8) = adj/hyp
-sqrt(5) / 3 = x/3
x = -sqrt(5)"}]},"trigonometric-functions-graphs-and-problems-i":{"title":"Trigonometric Functions Graphs and Problems I","sections":[{"text":"

Trigonometric Functions Graphs and Problems I

Alexander DeCarlo

While we are familiar with the six trigonometric functions through the unit circle, let\'s explore what these functions look like when graphed. Our x value will represent the angle value in radians, and the y value will be the corresponding return value.

Let\'s examine the special values in the sin function that are rational: 0 - 0, \u03c0/2 - 1, \u03c0 - 0, 3\u03c0/2 - -1, and 2\u03c0 - 0. We observe this cycling motion every 2\u03c0 radians, known as the period. Let\'s visualize this graphically.

\\"Graph

Upon graphing and examination, a beautiful cycling pattern emerges.

Since sin(x) = cos(x - \u03c0/2), we anticipate a \u03c0/2 shift to the right for the cosine graph.

\\"Graph"},{"text":"

Now, let\'s apply standard function transformations to see how these functions change with applications such as f(x) = A sin[B(x - h)] + k. This applies to both sin and cos.

Let\'s examine each transformation individually:

When looking at sin and cos, which cycle between 1 and -1, we notice a change by a value from 1 in opposite directions from the average line, 0. Therefore, |A| is the amplitude, representing the distance the maximum and minimum values deviate from the midline point.

Note: A negative sign on A could flip everything across the x-axis. A = (max - min) / 2, as this finds the average distance from the midline.

B represents the period. In a normal wave, it repeats every 2\u03c0. When B increases, the entire thing compresses, resulting in a period of 2\u03c0 / B.

h denotes the horizontal shift, and k is the vertical shift. To find k, you can find the actual average for the midline: (max + min) / 2.

"},{"text":"

EXAMPLES:

f(x) = 3cos(2x) + 1
g(x) = 2sin(\u03c0x + \u03c0) - 3

For f(x):

k is 1, making the midline 1, and A is 3, so it will extend up and down by 3. The 2x means this will happen during the period \u03c0.

\\"Graph

For g(x):

A is 2. No reflection is needed. B is \u03c0, so the period is 2\u03c0 / \u03c0, which is just 2. x is shifted left by 1, and the midline is -3.

\\"Graph

What problems can we model with our knowledge of the graphs of sines and cosines?

Consider the word problem: A Ferris wheel with a diameter of 520 feet has a low point of 30 feet. Every 30 minutes, it completes 1 rotation (starting at the bottom). Let f(x) model the height of a point on the Ferris wheel.

We begin at the lowest point, and since a cos graph always begins at 1, a negative cos must start at the bottom.

If the diameter is 520 feet, then we go 260 feet up and down from the midline, meaning a = 260.

If a full rotation is 30 minutes, then 30 = 2\u03c0 / B. B = \u03c0 / 15.

To start at 30, instead of the current -260, we need to add 290.

So, f(x) = -260cos(\u03c0 / 15 * x) + 290.

"}]},"trigonometric-functions-graphs-and-problems-ii":{"title":"Trigonometric Functions Graphs and Problems II","sections":[{"text":"

Trigonometric Functions Graphs and Problems II

Alexander DeCarlo

We graphed cos and sin, but how would we graph their reciprocal functions, csc and sec? Remember that csc(x) = 1/sin(x) and sec(x) = 1/cos(x). This is very important. Realizing that sin and cos are in the denominator, we can also see that whenever they return zero, the fraction would be 1/0, which is undefined. These occur every \u03c0 period and are known as asymptotes.

\\"Graph

Here for the csc graph, these are all the points we know are undefined. What happens when we approach those points? They must either approach infinity or negative infinity, as when the denominator gets smaller, it returns a larger and larger value. When the sin graph is in the positive half of the coordinate plane, it must blow up to infinity, and when the sin graph is negative, it will expand to negative infinity. Let\'s use a sine graph to visualize this:

\\"Graph

As we see, some parts stay between 0 and 1 and others go from 0 to -1. Our csc graph will touch the sin graph when it peaks at 1 or -1, as 1/1 = 1, but will diverge from there. In the negative parts, to negative infinity. With this information, we can create our cosecant graph in orange.

The secant graph is done the exact same way. This process helps you graph any sec or csc graph, as you simply graph the sin or cos graph and then apply the asymptote concept.

tan(x) and cot(x) graphs:

tan(x) is undefined at \u03c0/2 + 2\u03c0k and 3\u03c0/2 + 2\u03c0k. Graphing tan is quite simple for this reason. We know when tan(x) goes negative and goes positive, so we can make a graph easily.

\\"Graph

With cot(x), we notice that the 0\'s will become the asymptotes, BUT the asymptotes in tan(x) will become the 0\'s for cot(x). These ideas are common to see with reciprocal functions.

\\"Graph

Comparing the two graphs, they almost look reflected. This comes from the identity that we previously derived that states tan(x + \u03c0/2) = -cot(x). Also, take note that for these graphs, the period is not 2\u03c0, just \u03c0.

y = 3cot(3\u03c0/2(x)).

Here we would have a stretch of 3, and a period of 3/2. Given that cot(x) has asymptotes at 0, \u03c0, 2\u03c0, etc.., we can solve for the asymptotes by taking both of these base asymptotes and then applying the proper transformations. :

3\u03c0/2(x) = 2\u03c0
x = 4/3

3\u03c0/2(x) = \u03c0
x = 2/3

We can see here that the steps in the asymptotes are in increments of 2/3, something we could also have calculated by knowing that the function repeats every \u03c0. Now, you could also calculate the 0\'s by directly applying transformations, but a much simpler way is to realize that the asymptotes of any trigonometric function are equidistant from the x intercept on the base function. This way, we can see that the 0\'s would be 1/3, 1, etc\u2026 Since this is a cot function, the function will continuously decrease. Given all of these baby steps calculated, try graphing this for yourself, since the challenging parts are out of the way."}]},"the-inverse-trigonometric-functions":{"title":"The Inverse Trigonometric Functions","sections":[{"text":"

The Inverse Trigonometric Functions

Alexander DeCarlo

The inverse trig functions, instead of returning ratios, give the angle that provides that ratio. The naming of these is quite easy, and simply just adds an \'arc\' to the functions. They can also be named with the inverse sign, what we are familiar with from the right and general angle trigonometry.

arccos(x) OR cos-1(x)
arcsin(x) OR sin-1(x)
arctan(x) OR tan-1(x)
arccsc(x) OR csc-1(x)
arcsec(x) OR sec-1(x)
arccot(x) OR cot-1(x)

For example, arcsin(1/2) = \u03c0/6

Why not also 5\u03c0/6? To make it a function, we have to restrict its domain to the right half of the unit circle, [-\u03c0/2, \u03c0/2]. arctan(x) is also only between [-\u03c0/2, \u03c0/2]. cos however must be defined as the top half of the unit circle,[0, \u03c0]. Similar to the square root, the domain restriction only exists when these functions are used alone, however in trigonometric equations, every angle will be allowed.

"},{"text":"

Composition

Consider sin(arccos(x)). How would we solve a composite function like this?

sin(arccos(x)) = \u221a(1 - x2)

We can resemble x/1 as a triangle since these are the lengths of the hypotenuse and adjacent.

\\"A

sin is opp / hyp. So, what is opp? x2 + opp2 = 1. opp = \u221a(1 - x2)

Therefore, sin(arccos(x)) = \u221a(1 - x2)

We can also notice the same process would work for: cos(arcsin(x)).

\u2234 sin(arccos(x)) = \u221a(1 - x2)
cos(arcsin(x)) = \u221a(1 - x2)

What about arccos(sin(x))?

sin(x) = cos(\u03c0/2 - x)

arccos(cos(\u03c0/2 - x))

The arccos and cos would cancel out, leaving \u03c0/2 - x.

\u2234 arccos(sin(x)) = \u03c0/2 - x
arcsin(cos(x)) = \u03c0/2 - x

"}]},"introduction-to-trigonometric-equations":{"title":"Introduction to Trigonometric Equations","sections":[{"text":"

Introduction to Trigonometric Equations

Alexander DeCarlo

With trigonometric equations, domain restrictions no longer apply; we consider every possible angle unless specified otherwise. For instance, 3\u03c0 / 4 + 2\u03c0k indicates solutions at 3\u03c0 / 4 and all angles obtained by adding or subtracting multiples of 2\u03c0, where k belongs to the set of integers (k \u220a \u2124).

"},{"text":"

2cos(\u1340) = sqrt(3)
cos(\u1340) = sqrt(3) / 2
\u1340 = \u03c0 / 6, but can also be 11\u03c0 / 6 because of the unit circle. You can also go fully 2\u03c0 around the circle, allowing the following (in all these cases k \u220a \u2124).
\u1340 = \u03c0 / 6 + 2\u03c0k, 11\u03c0 / 6 + 2\u03c0k

2sin(\u1340) = -1
sin(\u1340) = -1 / 2
\u1340 = 7\u03c0 / 6 + 2\u03c0k, 11\u03c0 / 6 + 2\u03c0k

Let\'s look at something more challenging:

5cos(3\u1340 + 1) = 5
cos(3\u1340 + 1) = 1

Let\'s substitute u for 3\u1340 + 1 to make this simpler.

cos(u) = 1
u = 2\u03c0k
3\u1340 + 1 = 2\u03c0k
\u1340 = (2\u03c0k - 1) / 3

2cos(2\u1340) = -sqrt(2)
cos(2\u1340) = -sqrt(2) / 2
cos(u) = -sqrt(2) / 2
u = 5\u03c0 / 2 + 2\u03c0k, 3\u03c0 / 2 + 2\u03c0k
2\u1340 = 5\u03c0 / 4 + 2\u03c0k
2\u1340 = 3\u03c0 / 4 + 2\u03c0k
\u1340 = 5\u03c0 / 8 + \u03c0k, 3\u03c0 / 8 + \u03c0k"},{"text":"

sin(x) = -0.29
arcsin(-0.29) = x
-0.294 \u2248 x, convert this to positive radians by subtracting it from 2\u03c0.
x \u2248 5.989

So how do we find the second angle? Let\'s look at the unit circle.

\\"A

Blue is the 5.989 radians. To get to its equivalent red angle, we can observe adding 0.294 to \u03c0 to get there.

x \u2248 3.436
x \u2248 3.436 + 2\u03c0k, 5.989 + 2\u03c0k

Let\'s combine the ideas with this equation.

7cos(3\u1340) = 4 on [0, 2\u03c0]
cos(3\u1340) = 4 / 7
u = 3\u1340
cos(u) = 4 / 7
u = arccos(4 / 7) *calc
u = 0.963
u = 2\u03c0 - 0.963
u = 5.320
3\u1340 = 5.320 + 2\u03c0k, 3\u1340 = .963 + 2\u03c0k
\u1340 = 1.773 + 2\u03c0k / 3, 0.321 + 2\u03c0k / 3

Since this is on [0, 2\u03c0], we can account for 3 revolutions, as a single revolution is 2\u03c0k / 3.

\u1340 = 0.961, 7.246, 13.529, 5.320, 11.603, 17.887

Let\'s solve our final introductory equation.

65 = -13cos(\u03c0 / 12 * (\u1340 - 5)) + 61
4 = -13cos(\u03c0 / 12 * (\u1340 - 5))
-4 / 13 = cos(\u03c0 / 12 * (\u1340 - 5))
-4 / 13 = cos(u)
arccos(-4 / 13) = u
1.884 = u
u = 4.399
\u03c0 / 12 * (\u1340 - 5) = 1.884 + 2\u03c0k
\u03c0 / 12 * (\u1340 - 5) = 4.399 + 2\u03c0k
\u1340 - 5 = 7.196 + 24k
\u1340 - 5 = 16.803 + 24k
\u1340 = 12.196 + 24k, 21.803 + 24k"}]},"factoring-trigonometric-equations":{"title":"Factoring Trigonometric Equations","sections":[{"text":"

Factoring Trigonometric Equations

Alexander DeCarlo

Let\'s review the identities we\'ve established, focusing on our Pythagorean identities:

sin2(x) + cos2(x) = 1
tan2(x) + 1 = sec2(x)
1 + cot2(x) = csc2(x)

These become useful when solving trigonometric equations through factoring. Now, let\'s explore a simple example.

2sin2(x) + sin(x) = 0
sin(x)(2sin(x) + 1) = 0

This gives two possible solutions:

sin(x) = 0
sin(x) = -\xbd

Which leads to:

x = \u03c0k, 11\u03c0/6 + 2\u03c0k, 7\u03c0/6 + 2\u03c0k.

"},{"text":"

These problems share similarities with quadratics. Now, let\'s tackle a slightly more challenging equation:

2sin2(x) - cos(x) = 1.

Here, we leverage the identity sin2(x) + cos2(x) = 1 by expressing sin2(x) in terms of cos(x):

sin2(x) = 1 - cos2(x)

Substitute this into the original equation:

2(1 - cos2(x)) - cos(x) = 1

Simplify to get:

-2cos2(x) - cos(x) = -1

Rearrange and factor like a quadratic:

-2cos2(x) - cos(x) + 1 = 0
2cos2(x) + cos(x) - 1 = 0

Factor further:

(2cos(x) - 1)(cos(x) + 1) = 0

This gives two solutions:

cos(x) = 1/2
cos(x) = -1

Therefore, x can be expressed as:

x = \u03c0/3 + 2\u03c0k, 5\u03c0/3 + 2\u03c0k, \u03c0 + 2\u03c0k

"}]},"fundamental-trigonometric-identities-i":{"title":"Fundamental Trigonometric Identities I","sections":[{"text":"

Fundamental Trigonometric Identities I

Alexander DeCarlo

Let\'s consider the unit circle with two points, P and Q. Q is represented by angle \u03b2, and P by \u03b1.

\\"A

Here, Q = (cos(\u03b2), sin(\u03b2)), and P = (cos(\u03b1), sin(\u03b1)).

\\"A

Now, P\' = (cos(\u03b1 - \u03b2), sin(\u03b1 - \u03b2)), and Q\' = (1, 0).

A crucial observation is that, since we rotated the triangle, the line PQ is equivalent to P\'Q\'.

We can set up the distances using the distance formula:

PQ = sqrt((cos(\u03b1) - cos(\u03b2))2 + (sin(\u03b1) - sin(\u03b2))2)
P\'Q\' = sqrt((cos(\u03b1 - \u03b2) - 1)2 + (sin(\u03b1 - \u03b2) + 0)2)

Setting them equal to each other:

(cos(\u03b1) - cos(\u03b2))2 + (sin(\u03b1) - sin(\u03b2))2 = (cos(\u03b1 - \u03b2) - 1)2 + sin2(\u03b1 - \u03b2)

Simplifying the equation:

2 - 2cos(\u03b1)cos(\u03b2) - 2sin(\u03b1)sin(\u03b2) = (cos(\u03b1 - \u03b2) - 1)2 + sin2(\u03b1 - \u03b2)
-2(-1 + cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2)) = -2(cos(\u03b1 - \u03b2) - 1)
-1 + cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2) = cos(\u03b1 - \u03b2) - 1

This derivation yields the formula: cos(\u03b1 - \u03b2) = cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2), a valuable identity in sum and difference identities.

Now, let\'s explore other identities by setting \u03b2 = -\u03b3:

cos(\u03b1 + \u03b3) = cos(\u03b1)cos(\u03b3) - sin(\u03b1)sin(\u03b3)

Thus, we obtain:

cos(\u03b1 + \u03b2) = cos(\u03b1)cos(\u03b2) - sin(\u03b1)sin(\u03b2)
cos(\u03b1 - \u03b2) = cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2)

These are the cosine sum and difference identities.

For sine, we can use the substitution trick for sin(x) = cos(x - \u03c0/2) and -sin(x) = cos(x + \u03c0/2).

Let \u03b1 = \u03c0/2 - \u03b3:

sin(\u03b3 + \u03b2) = sin(\u03b3)cos(\u03b2) + cos(\u03b3)sin(\u03b2)
sin(\u03b3 - \u03b2) = sin(\u03b3)cos(\u03b2) - cos(\u03b3)sin(\u03b2)

Thus, we derive:

cos(\u03b1 + \u03b2) = cos(\u03b1)cos(\u03b2) - sin(\u03b1)sin(\u03b2)
cos(\u03b1 - \u03b2) = cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2)
sin(\u03b1 + \u03b2) = sin(\u03b1)cos(\u03b2) + cos(\u03b1)sin(\u03b2)
sin(\u03b1 - \u03b2) = sin(\u03b1)cos(\u03b2) - cos(\u03b1)sin(\u03b2)

These are our sum and difference identities, crucial for solving various trigonometric equations.

"}]},"fundamental-trigonometric-identities-ii":{"title":"Fundamental Trigonometric Identities II","sections":[{"text":"

Fundamental Trigonometric Identities II

Alexander DeCarlo

We previously discovered these identities:

cos(\u03b1 + \u03b2) = cos(\u03b1)cos(\u03b2) - sin(\u03b1)sin(\u03b2)
cos(\u03b1 - \u03b2) = cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2)
sin(\u03b1 + \u03b2) = sin(\u03b1)cos(\u03b2) - cos(\u03b1)sin(\u03b2)
sin(\u03b1 - \u03b2) = sin(\u03b1)cos(\u03b2) - cos(\u03b1)sin(\u03b2)

How can we use them to discover MORE identities? Well, it\'s quite simple! Notice that you can cancel out certain terms by adding equations:

cos(\u03b1 + \u03b2) = cos(\u03b1)cos(\u03b2) - sin(\u03b1)sin(\u03b2)
cos(\u03b1 - \u03b2) = cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2)

cos(\u03b1 + \u03b2) + cos(\u03b1 - \u03b2) = 2cos(\u03b1)cos(\u03b2)
(cos(\u03b1 + \u03b2) + cos(\u03b1 - \u03b2)) / 2 = cos(\u03b1)cos(\u03b2).

We can repeat this process and we can find identities for all pairs of multiplying sine and cosine, which gives us:

cos(\u03b1)cos(\u03b2) = (cos(\u03b1 + \u03b2) + cos(\u03b1 - \u03b2)) / 2
sin(\u03b1)sin(\u03b2) = (cos(\u03b1 - \u03b2) - cos(\u03b1 + \u03b2)) / 2
sin(\u03b1)cos(\u03b2) = (sin(\u03b1 + \u03b2) + sin(\u03b1 - \u03b2)) / 2

These are our products to sum identities.

The product to sum identities are particularly useful anytime we want to convert multiplication of trigonometric functions into addition and subtraction. This will become useful when we eventually put these new identities to use. Now that we know product -> sum identities, how would we go upon getting sum -> product identities? The key is in rearranging \u03b1 and \u03b2 in our product to sum identities.

Given: sin(\u03b1)cos(\u03b2) = (sin(\u03b1 + \u03b2) + sin(\u03b1 - \u03b2) / 2

let u = \u03b1 + \u03b2, v = \u03b1 - \u03b2. We must get u and v in terms of \u03b1 and \u03b2 so we can change the entire equation to be in terms of u and v.

v = \u03b1 - \u03b2
u = \u03b1 + \u03b2

Adding these,

v + u = 2\u03b1
(u+v) / 2 = \u03b1

Subtracting instead of adding,

v - u = -2\u03b2
(u - v) / 2 = \u03b2

Substituting into our given, we get:

sin((u+v) / 2)cos((u - v) / 2)) = (sin(u) + sin(v)) / 2
2sin((u+v) / 2 )cos((u - v) / 2)) = sin(u) + sin(v)

If we let w = -v, we can find the subtraction variant.

2sin((u - w) / 2)cos((u + w) / 2)) = sin(u) - sin(w)

Using the exact same process, we can find these identities for sine and cosine, which are as follows:

2cos((u + v) / 2))cos((u - v) / 2)) = cos(u) + cos(v)
2sin((u + v) / 2))sin((u - v) / 2)) = cos(u) - cos(v)

\u2234
2sin((u + v) / 2 )cos((u - v) / 2 )) = sin(u) + sin(v)
2sin((u - w) / 2 )cos((u + w) / 2 )) = sin(u) - sin(w)
2cos((u + v) / 2))cos((u - v) / 2)) = cos(u) + cos(v)
2sin((u + v) / 2))sin((u - v) / 2)) = cos(u) - cos(v)

"}]},"using-the-sum-and-difference-identity":{"title":"Using the Sum and Difference Identity","sections":[{"text":"

Using the Sum and Difference Identity

Alexander DeCarlo

The sum and difference identities, the most fundamental of the more complex trigonometric identities, can be used in many ways, including finding exact values to previously unknown angles.

Let\'s consider the sum identity for sin(x).

sin(\u03b1 + \u03b2) = sin(\u03b1) * cos(\u03b2) + sin(\u03b2) * cos(\u03b1)

We can find unknown angle values by adding or subtracting known ones. For example, let\'s consider \u03c0/6 and \u03c0/4. Summing these together, we get 5\u03c0/12, an angle that we currently have no clue what the exact value is.

sin(5\u03c0/12) = sin(\u03c0/4) * cos(\u03c0/6) + sin(\u03c0/6) * cos(5\u03c0/12)
= (1/2) * (sqrt(2) / 2) + sqrt(3) / 2 * sqrt(2) / 2
= (sqrt(2) + sqrt(6)) / 4

What about a base fraction, like \u03c0/12? Knowing the exact value of that could help us find the exact value of a lot more angles. Here, we can use the difference formula.

What subtracts to equal \u03c0/12? \u03c0/3 and \u03c0/4 do!

sin(\u03b1 - \u03b2) = sin(\u03b1) * cos(\u03b2) - sin(\u03b2) * cos(\u03b1)
sin(\u03c0/12) = sin(\u03c0/3) * cos(\u03c0/4) - sin(\u03c0/3) * cos(\u03c0/4)
sin(\u03c0/12) = (sqrt(6) - sqrt(2)) / 4

Certainly isn\'t as nice looking as something like sin(\u03c0/3), which is why these values usually do not need to be memorized. This all goes to show that any trigonometric function will yield an exact value if the angle is in terms of \u03c0, due to the transcendentality of the trigonometric functions and the transcendentality of \u03c0.

sin(5 radians) will never be in exact value (it would be transcendental), but sin(3\u03c0/19) does have an exact value.

"},{"text":"

Simplification of the Sum and Difference Identities

Oftentimes, we find that the sum and difference identities are a bit too general for what we need to deal with. These can simplify to what are known as the double angle identities.

sin(\u03b1 + \u03b2) = sin(\u03b1) * cos(\u03b2) + sin(\u03b2) * cos(\u03b1)

What if \u03b1 = \u03b2?

sin(2\u03b1) = sin(\u03b1) * cos(\u03b1) + sin(\u03b1) * cos(\u03b1)
sin(2\u03b1) = 2sin(\u03b1) * cos(\u03b1)

Also note that cos(\u03b1 + \u03b2) = cos(\u03b1) * cos(\u03b2) - sin(\u03b1) * sin(\u03b2).

cos(2\u03b1) = cos(\u03b1) * cos(\u03b1) - sin(\u03b2) * sin(\u03b2)
cos(2\u03b1) = cos2(\u03b1) - sin2(\u03b1)

These can make it easier to expand out something like sin(3\u03b1).

sin(3\u03b1) = sin(2\u03b1 + \u03b1)
sin(2\u03b1 + \u03b1) = sin(2\u03b1) * cos(\u03b1) + sin(\u03b1) * cos(2\u03b1)
= 2sin(\u03b1) * cos(\u03b1) * cos(\u03b1) + sin(\u03b1) * (cos2(\u03b1) - sin2(\u03b1))
= 2sin(\u03b1) * cos2(\u03b1) + sin(\u03b1) * cos2(\u03b1) - sin(\u03b1) * sin2(\u03b1)
= 3sin(\u03b1) * cos2(\u03b1) - sin3(\u03b1)
= 3sin(\u03b1) * cos2(\u03b1) - sin3(\u03b1)
= 3sin(\u03b1) * (1-sin^2(\u03b1)) - sin3(\u03b1)
= 3sin(\u03b1) - 3sin3(\u03b1) - sin3(\u03b1)
sin(3\u03b1) = 3sin(\u03b1) - 4sin3(\u03b1)

Notice that there may be a relationship between the overall degree of the relationship and the inner coefficient of sin(3\u03b1). We will dive deeper into this in later articles.

"}]},"using-identities-single-sine-expression":{"title":"Using Identities: Single Sine Expression","sections":[{"text":"

Using Identities: Single Sine Expression

Alexander DeCarlo

As we journey through mathematics, we may find it very useful to rewrite trigonometric expressions as sine functions. This becomes helpful when solving trigonometric equations in some cases, but also with calculus and graphing. Let\'s see what we can do with a general sine expression.

The general form is A * sin(Bx + C) where A, B and C are constants. Let\'s try to separate this out. We want to work backwards from the answer because we are trying to convert complex expressions to this general form.

Using the sine addition identity, we can expand to A * [sin(Bx) * cos(C) + cos(Bx) * sin(C)]. Notice that sin(C) and cos(C) are constants. Carrying on, let\'s distribute out constant A and move the constants in front.

A * cos(C)sin(Bx) + A * sin(C)cos(Bx)

Since A * cos(C) and A * sin(B) are constant values, let\'s rewrite them as M and N respectively.

M * sin(Bx) + N * cos(Bx)
M = A * cos(C)
N = A * sin(C)

Hold on a second. We can use one of our identities to find a relationship between M, N, and A to remove those nasty sine and cosine functions.

M2 + N2 = (A * cos(C))2 + (A * sin(C))2
M2 + N2 = A2(cos2(C) + sin2(C))
M2 + N2 = A2

We have two more trivial relationships to uncover:

M = A * cos(C)
N = A * sin(C)

This means M / A = cos(C) and N / A = sin(C). Let\'s see what we managed to discover here.

M * sin(Bx) + N * cos(Bx) = A * sin(Bx + C)
M2 + N2 = A2
M / A = cos(C)
N / A = sin(C)

We can now turn any expression that involves adding cosine and sine with the same angle into a single sine function to be solved or graphed!

Let\'s look at the most simple of these expressions: cos(x) + sin(x).

M = 1
N = 1

(1)2 + (1)2 = A2

A = sqrt(2)
B = 1, since x * 1 = x in cos(x) and sin(x)

cos(C) = 1 / sqrt(2) = sqrt(2) / 2
C = \u03c0/4, 7\u03c0/4

sin(C) = sqrt(2) / 2
C = \u03c0/4, 3\u03c0/4

Here, we see that \u03c0/4 satisfies both of these relationships, so cos(x) + sin(x) = sqrt(2) * sin(x + \u03c0/4).

Let\'s try another!

4sqrt(3) * sin(5x) - 4cos(5x)

M = 4sqrt(3)
N = -4

16 + 48 = A2
A = 8

cos(C) = 4sqrt(3) / 8
sin(C) = -4/8

The value that satisfies both equations is 11\u03c0/6. Thus, 4sqrt(3) * sin(5x) - 4cos(5x) = 8sin(5x + 11\u03c0/6).

"}]},"solving-trigonometric-equations-ii":{"title":"Solving Trigonometric Equations II","sections":[{"text":"

Solving Trigonometric Equations II

Now that we have derived some identities and found some tricks within trigonometry, let\'s solve some equations that we could not previously solve before. Let\'s start with a problem involving sine and cosine.

2sin(4x) - 8cos(4x) = sin(2x) * cos(2x) + 3

This looks challenging, but it\'s quite simple with our knowledge of identities. First, we can use the single sine trick, then the product to sum identity. We could also multiply both sides of this equation by 2, and use the double angle identity. Both work here, but generally try to find the fastest solution for when things get harder.

4sin(4x) - 16cos(4x) = 2sin(2x) * cos(2x) + 6
4sin(4x) - 16cos(4x) = sin(4x) + 6
3sin(4x) - 16cos(4x) = 6

Now we can turn this into a single sine.

32 + 162 = A2

A = sqrt(265)
B = 4

3 / sqrt(265) = sin(C)
16 / sqrt(265) = cos(C)

C = 0.185 rad, where both equations are true.

sqrt(265) * sin(4x + 0.185) = 6
sin(4x + .185) = 6 / sqrt(265)

Remember, treat everything within sin() as u.

sin(u) = 6 / sqrt(265)

u = 0.377 + 2\u03c0n, where n is an integer.
u = 2.764 + 2\u03c0n

Substitute in for u.

4x + 0.185 = 0.377 + 2\u03c0n
4x = 0.192 + 2\u03c0n
x = 0.048 + \u03c0n / 2

4x + 0.185 = 2.764 + 2\u03c0n
x = 0.720 + \u03c0n / 2

Let\'s solve another, simpler equation for some exact values.

sin(x) * sin(2x) + cos(x) * cos(2x) = sqrt(3) / 2

Here, use the difference formula for cosine. When solving for equations, we should condense without losing a possible root.

cos(x - 2x) = sqrt(3) / 2
cos(-x) = sqrt(3) / 2

Remember that cosine isn\'t affected by that negative value, you can memorize this for efficiency, but it never hurts to sketch the unit circle to see this simple relationship.

cos(x) = sqrt(3) / 2

Here x = \u03c0/6, 11\u03c0/6.

Let\'s solve one more.

sin(x) + sin(3x) = cos(x) [0, 2\u03c0]

We can\'t use previous methods in this article, but we luck out because there isn\'t a constant term. This means if we can get a product form and factor, we can solve it. Good thing there\'s an identity for that!

2sin((x + 3x) / 2) * cos((x - 3x) / 2) = cos(x)
2sin(2x) * cos(-x) = cos(x)
2sin(2x) * cos(x) = cos(x)
2sin(2x) * cos(x) - cos(x) = 0
cos(x) * [2sin(2x) - 1] = 0
cos(x) = 0

x = \u03c0/2, 3\u03c0/2

sin(2x) = 1/2
2x = \u03c0/6, 5\u03c0/6

x = \u03c0/12, 5\u03c0/12 (add \u03c0 to navigate through the interval)
x = 13\u03c0/12, 17\u03c0/12

In the end, x = \u03c0/2, 3\u03c0/2, \u03c0/12, 5\u03c0/12, 13\u03c0/12, 17\u03c0/12.

"}]}}')}}]); \ No newline at end of file +"use strict";(self.webpackChunkmathinfo=self.webpackChunkmathinfo||[]).push([[404],{404:function(s){s.exports=JSON.parse('{"Summary":{"text":"

Trigonometry

","image":"
\\"\\"

Trigonometry graphic.

"},"right-triangle-trigonometry":{"title":"Right Triangle Trigonometry","sections":[{"text":"

Right Triangle Trigonometry

Alexander DeCarlo

Trigonometry, as in the name, deals with the study of triangles, angles, ratios, trigonometric functions, graphs and a lot more. It is one of the most important topics in Mathematics and is the gateway for numerous other advanced concepts. While math revolves around Trigonometry, Trigonometry revolves around right triangles. A major reason for this is that only right triangles satisfy the Pythagorean Theorem, which states a2 + b2 = c2, where sides a and b are the legs, while side c is the hypotenuse. One of the key aspects of right triangles is their strict vertical and horizontal side due to the 90\xb0 angle. This also means side c is essentially the slope (rise / run) of the triangle. Let\'s take a look at this right triangle:

\'A","image":"
\\"\\"

Diagram of right triangle.

"},{"text":"

With just this diagram, we can identify many different ratios between the sides. There are a total of 6 ratios, each a trigonometric function. These operations are designated as functions because they completely rely on the angle between both sides.

Let\'s focus on angle \u03b1. Starting with the sine function, we know sin = opposite / hypotenuse. Applying that to angle \u03b1, we find that sin(\u03b1) = a / c, because side a is opposite to angle \u03b1, and side c is the hypotenuse of the triangle. Sine works the exact same way with any other angle. For example, sin(\u03b2) = b / c, as side b is opposite to angle \u03b2, and side c is again the hypotenuse of the triangle. Now that we know all about sine, let\'s look at the other core trigonometric functions:

\u03b8 for any given angle:
Sine -> sin(\u03b8) = opposite / hypotenuse
Cosine -> cos(\u03b8) = adjacent / hypotenuse
Tangent -> tan(\u03b8) = opposite / adjacent

Swapping the numerator and denominator of these core functions gives us their reciprocals:



Cosecant -> csc(\u03b8) = hypotenuse / opposite = 1 / sin(\u03b8)
Secant -> sec(\u03b8) = adjacent / hypotenuse = 1 / cos(\u03b8)
Cotangent -> cot(\u03b8) = adjacent / opposite = 1 / tan(\u03b8)

These 6 functions are all extremely important, with relationships to be further elucidated on as we go through trigonometry. For now though, it\'s important to simply understand where these functions come from and how they\'re used in relation to right triangles.

","image":"
\\"\\"

All 6 trigonometric functions of an angle on a triangle of side lengths 5, 4, and 3.

"},{"text":"\'A

Trigonometry will help us solve triangles like these. Our goal is to find side length c, but we aren\'t given both sides a and b. So what is side c? Which trigonometric function will encompass both side length11 and side c to return the angle? Looking at the 42\xb0 angle, we can see the side with length 11 is adjacent to it, and side c is the hypotenuse of the triangle. Going back to the trigonometric functions, cosine conveniently uses both the adjacent and hypotenuse sides of a right triangle.



cos(42\xb0) = 11 / c
c * cos(42\xb0) = 11
c = 11 / cos(42\xb0)

Here cos(42\xb0) is just a constant, as cos(42\xb0) will be the same for all triangles with a 42\xb0 angle. Using a calculator equipped with the cosine function, we arrive at c \u2248 14.8.

","image":"
\\"\\"

Can you find the length of side b? (b \u2248 6.3)

"}]},"right-angle-trigonometry-test-problem":{"title":"Right Angle Trigonometry Test Problem","sections":[{"text":"

Right Angle Trigonometry Test Problem

Alexander DeCarlo\\"A

Solution:Let y = the base of the entire triangle.
tan(30\xb0) = 50 / y
y * tan(30\xb0) = 50
y = 50 / tan(30\xb0)

Let z = the base of the inner right triangle, z = y - x.
tan(60\xb0) = 50 / z
z * tan(60\xb0) = 50
z * tan(60\xb0) = 50

x = y - z
x = 50 / tan(30\xb0) - 50 / tan(60\xb0)
x \u2248 57.7

","image":"
\\"\\"

All 6 trigonometric functions of an angle on a triangle of side lengths 5, 4, and 3.

"}]},"general-angle-trigonometry-i":{"title":"General Angle Trigonometry I","sections":[{"text":"

General Angle Trigonometry I

Alexander DeCarlo

How can you extend the use of trigonometry to non-right angle triangles with trigonometric functions? The key is in breaking up the \u201cGeneral Triangle\u201d into right triangles and seeing what can be done from there. Let\'s look at the general triangle, where every angle, not just two, are unknown.

\\"Standard

By dropping an altitude at the height to the base of the general triangle, we have created two right triangles. Looking more closely, we can observe a relationship between the two triangles.

sin(B) = h / a
sin(A) = h / b

Then, we can solve for the height.

a * sin(B) = h
b * sin(A) = h

Taking advantage of their equality, we can merge the equations into the following:

a * sin(B) = b * sin(A)

And finally, dividing both sides by ab yields us our final result.

sin(B) / b = sin(A) / a

Realizing that we placed the altitude at an arbitrary angle, we could have rotated the triangle and placed it anywhere else for the same relationship. Therefore, sin(C) / c must also be equivalent.

\u2234 sin(A) / a = sin(B) / b = sin(C) / c

This very simple proof establishes the theorem known as The Law of Sines.

So what kinds of triangles can we solve with this?

"},{"text":"

AAS Triangles

Solving triangles refers to finding every angle and side length associated with that triangle. AAS - Angle Angle Side, means you know two angles and a side, with the order of appearance being angle-angle-side. Because the direction of rotation is arbitrary to its properties, SAA is functionally the same thing. ASA means you know two angles and a side in between. Looping back to AAS/SAA triangles, they are the easiest to solve.

\\"Standard

To find missing angle C in the above description, simply subtract the known angles by 180\xb0 as all triangle degrees add to 180\xb0. C = 110. Finding the other sides is simply a matter of using The Law of Sines.

sin(A) / a = sin(B) / b = sin(C) / c, meaning the angle maintains a ratio with its opposite side, enabling us to find a missing angle/side assuming we have its counterpart and a full angle/side pair.

sin(30\xb0) / a = sin(40\xb0) / 11
a * sin (40\xb0) = 11sin(30\xb0)
a = 11sin(30\xb0) / sin(40\xb0), a \u2248 8.56

Solving for angle C would follow the same process.

"},{"text":"

SSA Triangles

SSA Triangles are more tricky to solve, as they lead to ambiguity which could result in two possible ways to create the triangle.

\\"Standard

We can set up:

sin(70\xb0) / 10 = sin(B) / 9
9sin(70\xb0) = 10sin(B)
sin(B) = 9sin(70\xb0) / 10

Here, we must use the inverse sine function. Basically, if sin(\u1340) gives us a side length, then sin-1(side length) gives us our angle. The only thing is, two possible angles satisfy the sine inverse in all cases. When you plug sin-1(9sin(70\xb0) / 10) into the calculator, you will get one value, the PRINCIPLE value. An explanation for all of this will be in later articles regarding trigonometric functions. sin-1(9sin(70\xb0) / 10) \u2248 57.7. Our other possible angle is simply 180 - PRINCIPLE ANGLE, in this case, 180 - 57.7 = 122.3. (For those who understand the unit circle or are revisiting from a later article, we are essentially reaching the equivalent sin(\u1340) value from QII or QIII.)

To see if both of these triangles are possible, let\'s try to get the 3rd value and see if it all makes sense.

180 - 57.7 - 70 = 52.3
180 - 122.3 - 70 = -12.3

We cannot have negative angles, so we can reject the 122.3\xb0 triangle. If it yielded a positive third angle, we would have to solve for the rest of our values for BOTH the triangles, and simply have two possibilities.

"}]},"general-angle-trigonometry-ii":{"title":"General Angle Trigonometry II","sections":[{"text":"

General Angle Trigonometry II

Alexander DeCarlo

The Law of Sines is an important relationship in the general triangle, but there\'s another core relationship we can implement when dealing with triangles, the Pythagorean Theorem. Consider the general triangle with an altitude dropped, creating two right triangles composing the general triangle. From side c, let\'s call x the base length of the smaller triangle and c - x the base length of the larger inner triangle.

\\"Standard

Instead of relying on sine, let\'s use cosine on angle B to try to do something with that x variable.

cos(B) = x / a
a * cos(B) = x

Now we have x in terms of our side lengths and angles of our general triangle. As we can see, we have two right triangles, so we can create some relationships using the Pythagorean Theorem. Note that we want to find an end relationship that doesn\'t contain variables such as h or x, as these aren\'t the side lengths or angles of our triangle.

h2 + x2 = a2
h2 = a2 - x2

h2 + (c - x)2 = b2
h2 = b2 - (c - x)2

To remove h, we can set these two equations equal to one another.

a2 - x2 = b2 - (c2 - 2cx + x2)
a2 - x2 = b2 - c2 + 2cx - x2
a2 = b2 - c2 + 2cx
a2 + c2 = b2 + 2cx

Remember, x = a * cos(B) so to remove x, substitute this in.

a2 + c2 - 2cb * cos(B) = b2

Since this is a general triangle, variable names can be changed and swapped as they are arbitrary, the only thing that has to stay is the relationship between the cosine angle and the side.

Therefore, to make it look better, we can change this to: a2 + b2 - 2ab * cos(\u03b3) = c2. The cosine angle stays opposite to side c, with the previous case opposite to side b.

This is called The Law of Cosines, for its use of the cos function instead of the sin function.

If you notice, The Law of Cosines is actually just an extension of the Pythagorean Theorem! When cos is 90\xb0, cos(90\xb0) = 0. This removes the -2ab term and simply gives us a2 + b2 = c2, which is pretty much a special case of The Law of Cosines.

"},{"text":"

SAS triangles

In this case, The Law of Sines is unusable, leaving us to rely on The Law of Cosines.

\\"Standard

We can find the missing side with The Law of Cosines.

92 + 82 - 2(9)(8)cos(40\xb0) = c2
34.7 = c2
c \u2248 5.9

We know all three sides, so to find an angle we can simply use The Law of Sines. We do NOT have to worry about the possibility of there being two triangles because if you know the lengths of all three sides, there can be only one possibility for what the angles can be.

"},{"text":"

SSS triangles

\\"Standard

To solve SSS triangles, we can use The Law of Cosines. To eliminate ambiguity when solving, solve for the angle of the largest side first. This removes the possibility of a second triangle because solving from the longest side gives us the largest angle.

52 + 62 - 2(5)(6)cos(x) = 72
25 + 36 - 60cos(x) = 49
-60cos(x) = -12
cos-1(12/60) = x
x \u2248 78.5

Proceed with The Law of Sines to get the other angles.

"}]},"the-unit-circle-i":{"title":"The Unit Circle I","sections":[{"text":"

The Unit Circle I

Alexander DeCarlo

To truly understand trigonometry, one must comprehend how angles work, understanding the unit circle, its relation to angles, and a novel angle measurement called radians.

Angles reach a maximum of 360\xb0, completing a full rotation in circular motion. This range from 0 - 360\xb0 exists because 360 is a conveniently divisible number (by 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, etc.). While practical for everyday use, it lacks the mathematical precision we desire.

\\"Circle

Consider a circle with a radius of 1, making the circumference C = 2\u03c0. This equates a full 360-degree rotation to 2\u03c0 radians, a unit used in advanced trigonometry. To convert between radians and degrees, multiply by 180/\u03c0 or \u03c0/180, respectively.

Now, observe the unit circle on the coordinate plane with an angle traversing it:

\\"Circle

Two angles are measured, originating from the right x-axis and extending outward. The blue and red lines, termed terminal rays, gauge the angle size. Notably, the red terminal ray measures externally, while the blue one measures internally, following the counterclockwise motion. For example, angle a is around 135\xb0 or 3\u03c0/4 radians, while angle b is greater. For angles exceeding 360\xb0, they continue counterclockwise, such as 500\xb0 being equivalent to 140\xb0 after a full rotation. Negative angles rotate clockwise; -1\xb0 equals 359\xb0. This insight allows us to divide the unit circle into four quadrants.

\\"Circle

Every \u03c0/4 (90\xb0) defines a new quadrant, with x and y lines acting as quadrantal separators.

Now that we understand this improved angle system and its application around the unit circle, let\'s explore its significance and connection to right triangles. Unlike general right triangles, special triangles like the 45-45-90 and 30-60-90 offer exact angle and side measurements.

\\"Two

These special triangles, created using the Pythagorean theorem and specific base splits, provide exact values without trigonometry. Placing these triangles within the unit circle, with a hypotenuse of length 1, allows us to compress them by dividing by 2 and 3, respectively. This results in three triangles per quadrant: 45-45-90, 30-60-90, and 60-30-90. This process enables the mapping of coordinates for various angles on the unit circle, forming a helpful diagram for memorization.

\\"A

How does this relate to trigonometric functions? The crucial point lies in the fact that the hypotenuse of these triangles is always 1. Thus, sin(\u03b8) = opp, where \\"opp\\" is the y-coordinate, and cos(\u03b8) = x-coordinate. This insight leads to the realization that coordinates on the circle are simply (cos(\u03b8), sin(\u03b8)). For instance, with our special triangle values, sin(\u03c0/4) = sqrt(2)/2.

Examining the unit circle reveals essential insights into trigonometric functions:

  • Sine and cosine values range between -1 and 1.

  • Sin and cosine values are equal at two values: \u03c0/4, 5\u03c0/4

  • In QI and QII, sin is positive; in QIV and QI, cos is positive.

  • As sin increases, cos decreases, and vice versa. Two different angles return the same value, either in the top or bottom half for sin and left or right for cos (sin(\u03c0 - \u1340) = sin\u1340, cos(2\u03c0 - \u1340) = cos\u1340).

  • Finally, recognizing that cos(\u1340) = x and sin(\u1340) = y, tan(\u1340), the slope, is found as y/x, making tan(\u1340) = sin(\u1340)/cos(\u1340). Discovering these relationships is termed finding identities, a concept explored further in our trigonometry studies.

"}]},"the-unit-circle-ii":{"title":"The Unit Circle II","sections":[{"text":"

The Unit Circle II

Alexander DeCarlo

Let\'s explore some intriguing relationships between trigonometric functions using the unit circle while maintaining clarity and coherence. First, let\'s examine the coordinates of points on circles with radii greater or less than 1.

sin(x) = y * r
r * sin(x) = y
r * cos(x) = x

Simply multiply by r, the radius. The ratios are still the same.

(r * sin(x), r * cos(x))

Consider the unit circle.

\\"A

Here, four angles are presented with colored terminal rays. The red and blue rays are reflected across the y-axis, and the blue and yellow across the x-axis. If we examine cos(blue) and cos(yellow), they are the same. Yellow is simply the negative angle of blue; thus, cos(-x) = cos(x). With sine, sin(-x) = -sin(x). Examining the relationship between blue and red, or yellow and purple, we observe that sin(\u03c0 - x) = sin(x), and cos(\u03c0 - x) = -cos(x).

We also observe that cos(2\u03c0 - x) = cos(x), and sin(2\u03c0 - x) = -sin(x). Knowing that tan(x) is the slope, when cos or sin is negative, tan will be negative. Therefore, tan(x + \u03c0) = tan(x), tan(x - \u03c0) = -tan(x), and tan(-x) = -tan(x). These relationships are intuitive and can be quickly deduced from the unit circle.

Now, let\'s explore the relationship between sine and cosine.

\\"A

When rotating 90\xb0, cos(x), which is x, would be rotated completely vertically upward, becoming sine, and sine would become cosine. Starting from the triangle in the first quadrant, upon rotation, the cosine and sine values swap places, and the sine becomes negative. This process is repeated in the second triangle, yielding the following relationships:

sin(x + \u03c0/2) = cos(x)
cos(x + \u03c0/2) = -sin(x)
sin(x - \u03c0/2) = -cos(x)
cos(x - \u03c0/2) = sin(x)

If we examine the tangent values, the cosine and sines are swapped, and one value becomes negative. Adding a right angle, \u03c0/2, yields the negative reciprocal of the original. Therefore:

tan(x + \u03c0/2) = tan(x - \u03c0/2)

More succinctly:

tan(x + \u03c0/2) = -cot(x)

These relationships showcase the beauty of trigonometric functions.

"}]},"the-unit-circle-iii":{"title":"The Unit Circle III","sections":[{"text":"

The Unit Circle III

Alexander DeCarlo

While we are acquainted with the representations of cos and sin on the unit circle, it\'s essential to explore how all trigonometric functions manifest and derive identities from these visual connections.

\\"A

Acknowledging that cos and sin correspond to our x and y values, the question arises: how do we express tan as a length measure? Leveraging the unit circle\'s triangle with a hypotenuse of 1, we can establish a right angle from the (cos, sin) point.

\\"A

Upon closer inspection, we observe that tan represents that line, given that tan is opp/adj, and adj is 1. This elucidates the name \\"tan,\\" signifying its tangent relationship to the circle. Additionally, it visually explains why tan(\u03c0/2) and tan(\u03c0/4) are undefined, extending infinitely without intersecting the x-axis. To determine sec, representing adj/opp, we leverage the triangle formed with tan, realizing that sec must denote the hypotenuse of that triangle.

\\"A

Applying a similar process, we can ascertain cotangent.

\\"A

Similarly, the derivation of cosecant follows a comparable procedure.

Exploring the relationships between trigonometric functions through the Pythagorean theorem, we unveil crucial connections:

sin2(x) + cos2(x) = 1
tan2(x) + 1 = sec2(x)
cot2(x) + 1 = csc2(x)
sin2(x) + (sec(x) - cos(x))2 = tan2(x)
sec2(x) + csc2(x) = (tan(x) + cot(x))2

These relationships are commonly known as the Pythagorean Identities. Among them, the most significant is sin2(x) + cos2(x) = 1, a fundamental identity that merits in-depth study.

"}]},"using-the-pythagorean-identity-sine-and-cosine":{"title":"Using the Pythagorean Identity: Sine and Cosine","sections":[{"text":"

Using the Pythagorean Identity: Sine and Cosine

Alexander DeCarlo

Let\'s review the previously derived Pythagorean Identity: sin2(x) + cos2(x) = 1.

Given: sin(\u03b8) = 2/3, \u03b8 in QII

\\"A

22 + x2 = 32
4 + x2 = 9
x2 = 5
x = \xb1sqrt(5)

cos(\u03b8) = adj/hyp
cos(\u03b8) = \xb1sqrt(5) / 3

Cosine must be negative in QII, therefore we reject positive sqrt(5) / 3.

x = -sqrt(5)

While this represents one method of solving, leveraging the Pythagorean Identity provides a more streamlined approach.

sin2(\u03b8) + cos2(\u03b8) = 1
(2/3)2 + cos2(\u03b8) = 1
cos2(\u03b8) = 5/9
cos \u03b8 = \xb1sqrt(5) / 3

Reject positive sqrt(5) / 3.

cos(\u03b8) = -sqrt(5) / 3

Then we can get the adjacent side from how we calculate cosine:

cos(\u03b8) = adj/hyp
-sqrt(5) / 3 = x/3
x = -sqrt(5)"}]},"trigonometric-functions-graphs-and-problems-i":{"title":"Trigonometric Functions Graphs and Problems I","sections":[{"text":"

Trigonometric Functions Graphs and Problems I

Alexander DeCarlo

While we are familiar with the six trigonometric functions through the unit circle, let\'s explore what these functions look like when graphed. Our x value will represent the angle value in radians, and the y value will be the corresponding return value.

Let\'s examine the special values in the sin function that are rational: 0 - 0, \u03c0/2 - 1, \u03c0 - 0, 3\u03c0/2 - -1, and 2\u03c0 - 0. We observe this cycling motion every 2\u03c0 radians, known as the period. Let\'s visualize this graphically.

\\"Graph

Upon graphing and examination, a beautiful cycling pattern emerges.

Since sin(x) = cos(x - \u03c0/2), we anticipate a \u03c0/2 shift to the right for the cosine graph.

\\"Graph"},{"text":"

Now, let\'s apply standard function transformations to see how these functions change with applications such as f(x) = A sin[B(x - h)] + k. This applies to both sin and cos.

Let\'s examine each transformation individually:

When looking at sin and cos, which cycle between 1 and -1, we notice a change by a value from 1 in opposite directions from the average line, 0. Therefore, |A| is the amplitude, representing the distance the maximum and minimum values deviate from the midline point.

Note: A negative sign on A could flip everything across the x-axis. A = (max - min) / 2, as this finds the average distance from the midline.

B represents the period. In a normal wave, it repeats every 2\u03c0. When B increases, the entire thing compresses, resulting in a period of 2\u03c0 / B.

h denotes the horizontal shift, and k is the vertical shift. To find k, you can find the actual average for the midline: (max + min) / 2.

"},{"text":"

EXAMPLES:

f(x) = 3cos(2x) + 1
g(x) = 2sin(\u03c0x + \u03c0) - 3

For f(x):

k is 1, making the midline 1, and A is 3, so it will extend up and down by 3. The 2x means this will happen during the period \u03c0.

\\"Graph

For g(x):

A is 2. No reflection is needed. B is \u03c0, so the period is 2\u03c0 / \u03c0, which is just 2. x is shifted left by 1, and the midline is -3.

\\"Graph

What problems can we model with our knowledge of the graphs of sines and cosines?

Consider the word problem: A Ferris wheel with a diameter of 520 feet has a low point of 30 feet. Every 30 minutes, it completes 1 rotation (starting at the bottom). Let f(x) model the height of a point on the Ferris wheel.

We begin at the lowest point, and since a cos graph always begins at 1, a negative cos must start at the bottom.

If the diameter is 520 feet, then we go 260 feet up and down from the midline, meaning a = 260.

If a full rotation is 30 minutes, then 30 = 2\u03c0 / B. B = \u03c0 / 15.

To start at 30, instead of the current -260, we need to add 290.

So, f(x) = -260cos(\u03c0 / 15 * x) + 290.

"}]},"trigonometric-functions-graphs-and-problems-ii":{"title":"Trigonometric Functions Graphs and Problems II","sections":[{"text":"

Trigonometric Functions Graphs and Problems II

Alexander DeCarlo

We graphed cos and sin, but how would we graph their reciprocal functions, csc and sec? Remember that csc(x) = 1/sin(x) and sec(x) = 1/cos(x). This is very important. Realizing that sin and cos are in the denominator, we can also see that whenever they return zero, the fraction would be 1/0, which is undefined. These occur every \u03c0 period and are known as asymptotes.

\\"Graph

Here for the csc graph, these are all the points we know are undefined. What happens when we approach those points? They must either approach infinity or negative infinity, as when the denominator gets smaller, it returns a larger and larger value. When the sin graph is in the positive half of the coordinate plane, it must blow up to infinity, and when the sin graph is negative, it will expand to negative infinity. Let\'s use a sine graph to visualize this:

\\"Graph

As we see, some parts stay between 0 and 1 and others go from 0 to -1. Our csc graph will touch the sin graph when it peaks at 1 or -1, as 1/1 = 1, but will diverge from there. In the negative parts, to negative infinity. With this information, we can create our cosecant graph in orange.

The secant graph is done the exact same way. This process helps you graph any sec or csc graph, as you simply graph the sin or cos graph and then apply the asymptote concept.

tan(x) and cot(x) graphs:

tan(x) is undefined at \u03c0/2 + 2\u03c0k and 3\u03c0/2 + 2\u03c0k. Graphing tan is quite simple for this reason. We know when tan(x) goes negative and goes positive, so we can make a graph easily.

\\"Graph

With cot(x), we notice that the 0\'s will become the asymptotes, BUT the asymptotes in tan(x) will become the 0\'s for cot(x). These ideas are common to see with reciprocal functions.

\\"Graph

Comparing the two graphs, they almost look reflected. This comes from the identity that we previously derived that states tan(x + \u03c0/2) = -cot(x). Also, take note that for these graphs, the period is not 2\u03c0, just \u03c0.

y = 3cot(3\u03c0/2(x)).

Here we would have a stretch of 3, and a period of 3/2. Given that cot(x) has asymptotes at 0, \u03c0, 2\u03c0, etc.., we can solve for the asymptotes by taking both of these base asymptotes and then applying the proper transformations. :

3\u03c0/2(x) = 2\u03c0
x = 4/3

3\u03c0/2(x) = \u03c0
x = 2/3

We can see here that the steps in the asymptotes are in increments of 2/3, something we could also have calculated by knowing that the function repeats every \u03c0. Now, you could also calculate the 0\'s by directly applying transformations, but a much simpler way is to realize that the asymptotes of any trigonometric function are equidistant from the x intercept on the base function. This way, we can see that the 0\'s would be 1/3, 1, etc\u2026 Since this is a cot function, the function will continuously decrease. Given all of these baby steps calculated, try graphing this for yourself, since the challenging parts are out of the way."}]},"the-inverse-trigonometric-functions":{"title":"The Inverse Trigonometric Functions","sections":[{"text":"

The Inverse Trigonometric Functions

Alexander DeCarlo

The inverse trig functions, instead of returning ratios, give the angle that provides that ratio. The naming of these is quite easy, and simply just adds an \'arc\' to the functions. They can also be named with the inverse sign, what we are familiar with from the right and general angle trigonometry.

arccos(x) OR cos-1(x)
arcsin(x) OR sin-1(x)
arctan(x) OR tan-1(x)
arccsc(x) OR csc-1(x)
arcsec(x) OR sec-1(x)
arccot(x) OR cot-1(x)

For example, arcsin(1/2) = \u03c0/6

Why not also 5\u03c0/6? To make it a function, we have to restrict its domain to the right half of the unit circle, [-\u03c0/2, \u03c0/2]. arctan(x) is also only between [-\u03c0/2, \u03c0/2]. cos however must be defined as the top half of the unit circle,[0, \u03c0]. Similar to the square root, the domain restriction only exists when these functions are used alone, however in trigonometric equations, every angle will be allowed.

"},{"text":"

Composition

Consider sin(arccos(x)). How would we solve a composite function like this?

sin(arccos(x)) = \u221a(1 - x2)

We can resemble x/1 as a triangle since these are the lengths of the hypotenuse and adjacent.

\\"A

sin is opp / hyp. So, what is opp? x2 + opp2 = 1. opp = \u221a(1 - x2)

Therefore, sin(arccos(x)) = \u221a(1 - x2)

We can also notice the same process would work for: cos(arcsin(x)).

\u2234 sin(arccos(x)) = \u221a(1 - x2)
cos(arcsin(x)) = \u221a(1 - x2)

What about arccos(sin(x))?

sin(x) = cos(\u03c0/2 - x)

arccos(cos(\u03c0/2 - x))

The arccos and cos would cancel out, leaving \u03c0/2 - x.

\u2234 arccos(sin(x)) = \u03c0/2 - x
arcsin(cos(x)) = \u03c0/2 - x

"}]},"introduction-to-trigonometric-equations":{"title":"Introduction to Trigonometric Equations","sections":[{"text":"

Introduction to Trigonometric Equations

Alexander DeCarlo

With trigonometric equations, domain restrictions no longer apply; we consider every possible angle unless specified otherwise. For instance, 3\u03c0 / 4 + 2\u03c0k indicates solutions at 3\u03c0 / 4 and all angles obtained by adding or subtracting multiples of 2\u03c0, where k belongs to the set of integers (k \u220a \u2124).

"},{"text":"

2cos(\u1340) = sqrt(3)
cos(\u1340) = sqrt(3) / 2
\u1340 = \u03c0 / 6, but can also be 11\u03c0 / 6 because of the unit circle. You can also go fully 2\u03c0 around the circle, allowing the following (in all these cases k \u220a \u2124).
\u1340 = \u03c0 / 6 + 2\u03c0k, 11\u03c0 / 6 + 2\u03c0k

2sin(\u1340) = -1
sin(\u1340) = -1 / 2
\u1340 = 7\u03c0 / 6 + 2\u03c0k, 11\u03c0 / 6 + 2\u03c0k

Let\'s look at something more challenging:

5cos(3\u1340 + 1) = 5
cos(3\u1340 + 1) = 1

Let\'s substitute u for 3\u1340 + 1 to make this simpler.

cos(u) = 1
u = 2\u03c0k
3\u1340 + 1 = 2\u03c0k
\u1340 = (2\u03c0k - 1) / 3

2cos(2\u1340) = -sqrt(2)
cos(2\u1340) = -sqrt(2) / 2
cos(u) = -sqrt(2) / 2
u = 5\u03c0 / 2 + 2\u03c0k, 3\u03c0 / 2 + 2\u03c0k
2\u1340 = 5\u03c0 / 4 + 2\u03c0k
2\u1340 = 3\u03c0 / 4 + 2\u03c0k
\u1340 = 5\u03c0 / 8 + \u03c0k, 3\u03c0 / 8 + \u03c0k"},{"text":"

sin(x) = -0.29
arcsin(-0.29) = x
-0.294 \u2248 x, convert this to positive radians by subtracting it from 2\u03c0.
x \u2248 5.989

So how do we find the second angle? Let\'s look at the unit circle.

\\"A

Blue is the 5.989 radians. To get to its equivalent red angle, we can observe adding 0.294 to \u03c0 to get there.

x \u2248 3.436
x \u2248 3.436 + 2\u03c0k, 5.989 + 2\u03c0k

Let\'s combine the ideas with this equation.

7cos(3\u1340) = 4 on [0, 2\u03c0]
cos(3\u1340) = 4 / 7
u = 3\u1340
cos(u) = 4 / 7
u = arccos(4 / 7) *calc
u = 0.963
u = 2\u03c0 - 0.963
u = 5.320
3\u1340 = 5.320 + 2\u03c0k, 3\u1340 = .963 + 2\u03c0k
\u1340 = 1.773 + 2\u03c0k / 3, 0.321 + 2\u03c0k / 3

Since this is on [0, 2\u03c0], we can account for 3 revolutions, as a single revolution is 2\u03c0k / 3.

\u1340 = 0.961, 7.246, 13.529, 5.320, 11.603, 17.887

Let\'s solve our final introductory equation.

65 = -13cos(\u03c0 / 12 * (\u1340 - 5)) + 61
4 = -13cos(\u03c0 / 12 * (\u1340 - 5))
-4 / 13 = cos(\u03c0 / 12 * (\u1340 - 5))
-4 / 13 = cos(u)
arccos(-4 / 13) = u
1.884 = u
u = 4.399
\u03c0 / 12 * (\u1340 - 5) = 1.884 + 2\u03c0k
\u03c0 / 12 * (\u1340 - 5) = 4.399 + 2\u03c0k
\u1340 - 5 = 7.196 + 24k
\u1340 - 5 = 16.803 + 24k
\u1340 = 12.196 + 24k, 21.803 + 24k"}]},"factoring-trigonometric-equations":{"title":"Factoring Trigonometric Equations","sections":[{"text":"

Factoring Trigonometric Equations

Alexander DeCarlo

Let\'s review the identities we\'ve established, focusing on our Pythagorean identities:

sin2(x) + cos2(x) = 1
tan2(x) + 1 = sec2(x)
1 + cot2(x) = csc2(x)

These become useful when solving trigonometric equations through factoring. Now, let\'s explore a simple example.

2sin2(x) + sin(x) = 0
sin(x)(2sin(x) + 1) = 0

This gives two possible solutions:

sin(x) = 0
sin(x) = -\xbd

Which leads to:

x = \u03c0k, 11\u03c0/6 + 2\u03c0k, 7\u03c0/6 + 2\u03c0k.

"},{"text":"

These problems share similarities with quadratics. Now, let\'s tackle a slightly more challenging equation:

2sin2(x) - cos(x) = 1.

Here, we leverage the identity sin2(x) + cos2(x) = 1 by expressing sin2(x) in terms of cos(x):

sin2(x) = 1 - cos2(x)

Substitute this into the original equation:

2(1 - cos2(x)) - cos(x) = 1

Simplify to get:

-2cos2(x) - cos(x) = -1

Rearrange and factor like a quadratic:

-2cos2(x) - cos(x) + 1 = 0
2cos2(x) + cos(x) - 1 = 0

Factor further:

(2cos(x) - 1)(cos(x) + 1) = 0

This gives two solutions:

cos(x) = 1/2
cos(x) = -1

Therefore, x can be expressed as:

x = \u03c0/3 + 2\u03c0k, 5\u03c0/3 + 2\u03c0k, \u03c0 + 2\u03c0k

"}]},"fundamental-trigonometric-identities-i":{"title":"Fundamental Trigonometric Identities I","sections":[{"text":"

Fundamental Trigonometric Identities I

Alexander DeCarlo

Let\'s consider the unit circle with two points, P and Q. Q is represented by angle \u03b2, and P by \u03b1.

\\"A

Here, Q = (cos(\u03b2), sin(\u03b2)), and P = (cos(\u03b1), sin(\u03b1)).

\\"A

Now, P\' = (cos(\u03b1 - \u03b2), sin(\u03b1 - \u03b2)), and Q\' = (1, 0).

A crucial observation is that, since we rotated the triangle, the line PQ is equivalent to P\'Q\'.

We can set up the distances using the distance formula:

PQ = sqrt((cos(\u03b1) - cos(\u03b2))2 + (sin(\u03b1) - sin(\u03b2))2)
P\'Q\' = sqrt((cos(\u03b1 - \u03b2) - 1)2 + (sin(\u03b1 - \u03b2) + 0)2)

Setting them equal to each other:

(cos(\u03b1) - cos(\u03b2))2 + (sin(\u03b1) - sin(\u03b2))2 = (cos(\u03b1 - \u03b2) - 1)2 + sin2(\u03b1 - \u03b2)

Simplifying the equation:

2 - 2cos(\u03b1)cos(\u03b2) - 2sin(\u03b1)sin(\u03b2) = (cos(\u03b1 - \u03b2) - 1)2 + sin2(\u03b1 - \u03b2)
-2(-1 + cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2)) = -2(cos(\u03b1 - \u03b2) - 1)
-1 + cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2) = cos(\u03b1 - \u03b2) - 1

This derivation yields the formula: cos(\u03b1 - \u03b2) = cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2), a valuable identity in sum and difference identities.

Now, let\'s explore other identities by setting \u03b2 = -\u03b3:

cos(\u03b1 + \u03b3) = cos(\u03b1)cos(\u03b3) - sin(\u03b1)sin(\u03b3)

Thus, we obtain:

cos(\u03b1 + \u03b2) = cos(\u03b1)cos(\u03b2) - sin(\u03b1)sin(\u03b2)
cos(\u03b1 - \u03b2) = cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2)

These are the cosine sum and difference identities.

For sine, we can use the substitution trick for sin(x) = cos(x - \u03c0/2) and -sin(x) = cos(x + \u03c0/2).

Let \u03b1 = \u03c0/2 - \u03b3:

sin(\u03b3 + \u03b2) = sin(\u03b3)cos(\u03b2) + cos(\u03b3)sin(\u03b2)
sin(\u03b3 - \u03b2) = sin(\u03b3)cos(\u03b2) - cos(\u03b3)sin(\u03b2)

Thus, we derive:

cos(\u03b1 + \u03b2) = cos(\u03b1)cos(\u03b2) - sin(\u03b1)sin(\u03b2)
cos(\u03b1 - \u03b2) = cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2)
sin(\u03b1 + \u03b2) = sin(\u03b1)cos(\u03b2) + cos(\u03b1)sin(\u03b2)
sin(\u03b1 - \u03b2) = sin(\u03b1)cos(\u03b2) - cos(\u03b1)sin(\u03b2)

These are our sum and difference identities, crucial for solving various trigonometric equations.

"}]},"fundamental-trigonometric-identities-ii":{"title":"Fundamental Trigonometric Identities II","sections":[{"text":"

Fundamental Trigonometric Identities II

Alexander DeCarlo

We previously discovered these identities:

cos(\u03b1 + \u03b2) = cos(\u03b1)cos(\u03b2) - sin(\u03b1)sin(\u03b2)
cos(\u03b1 - \u03b2) = cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2)
sin(\u03b1 + \u03b2) = sin(\u03b1)cos(\u03b2) - cos(\u03b1)sin(\u03b2)
sin(\u03b1 - \u03b2) = sin(\u03b1)cos(\u03b2) - cos(\u03b1)sin(\u03b2)

How can we use them to discover MORE identities? Well, it\'s quite simple! Notice that you can cancel out certain terms by adding equations:

cos(\u03b1 + \u03b2) = cos(\u03b1)cos(\u03b2) - sin(\u03b1)sin(\u03b2)
cos(\u03b1 - \u03b2) = cos(\u03b1)cos(\u03b2) + sin(\u03b1)sin(\u03b2)

cos(\u03b1 + \u03b2) + cos(\u03b1 - \u03b2) = 2cos(\u03b1)cos(\u03b2)
(cos(\u03b1 + \u03b2) + cos(\u03b1 - \u03b2)) / 2 = cos(\u03b1)cos(\u03b2).

We can repeat this process and we can find identities for all pairs of multiplying sine and cosine, which gives us:

cos(\u03b1)cos(\u03b2) = (cos(\u03b1 + \u03b2) + cos(\u03b1 - \u03b2)) / 2
sin(\u03b1)sin(\u03b2) = (cos(\u03b1 - \u03b2) - cos(\u03b1 + \u03b2)) / 2
sin(\u03b1)cos(\u03b2) = (sin(\u03b1 + \u03b2) + sin(\u03b1 - \u03b2)) / 2

These are our products to sum identities.

The product to sum identities are particularly useful anytime we want to convert multiplication of trigonometric functions into addition and subtraction. This will become useful when we eventually put these new identities to use. Now that we know product -> sum identities, how would we go upon getting sum -> product identities? The key is in rearranging \u03b1 and \u03b2 in our product to sum identities.

Given: sin(\u03b1)cos(\u03b2) = (sin(\u03b1 + \u03b2) + sin(\u03b1 - \u03b2) / 2

let u = \u03b1 + \u03b2, v = \u03b1 - \u03b2. We must get u and v in terms of \u03b1 and \u03b2 so we can change the entire equation to be in terms of u and v.

v = \u03b1 - \u03b2
u = \u03b1 + \u03b2

Adding these,

v + u = 2\u03b1
(u+v) / 2 = \u03b1

Subtracting instead of adding,

v - u = -2\u03b2
(u - v) / 2 = \u03b2

Substituting into our given, we get:

sin((u+v) / 2)cos((u - v) / 2)) = (sin(u) + sin(v)) / 2
2sin((u+v) / 2 )cos((u - v) / 2)) = sin(u) + sin(v)

If we let w = -v, we can find the subtraction variant.

2sin((u - w) / 2)cos((u + w) / 2)) = sin(u) - sin(w)

Using the exact same process, we can find these identities for sine and cosine, which are as follows:

2cos((u + v) / 2))cos((u - v) / 2)) = cos(u) + cos(v)
2sin((u + v) / 2))sin((u - v) / 2)) = cos(u) - cos(v)

\u2234
2sin((u + v) / 2 )cos((u - v) / 2 )) = sin(u) + sin(v)
2sin((u - w) / 2 )cos((u + w) / 2 )) = sin(u) - sin(w)
2cos((u + v) / 2))cos((u - v) / 2)) = cos(u) + cos(v)
2sin((u + v) / 2))sin((u - v) / 2)) = cos(u) - cos(v)

"}]},"using-the-sum-and-difference-identity":{"title":"Using the Sum and Difference Identity","sections":[{"text":"

Using the Sum and Difference Identity

Alexander DeCarlo

The sum and difference identities, the most fundamental of the more complex trigonometric identities, can be used in many ways, including finding exact values to previously unknown angles.

Let\'s consider the sum identity for sin(x).

sin(\u03b1 + \u03b2) = sin(\u03b1) * cos(\u03b2) + sin(\u03b2) * cos(\u03b1)

We can find unknown angle values by adding or subtracting known ones. For example, let\'s consider \u03c0/6 and \u03c0/4. Summing these together, we get 5\u03c0/12, an angle that we currently have no clue what the exact value is.

sin(5\u03c0/12) = sin(\u03c0/4) * cos(\u03c0/6) + sin(\u03c0/6) * cos(5\u03c0/12)
= (1/2) * (sqrt(2) / 2) + sqrt(3) / 2 * sqrt(2) / 2
= (sqrt(2) + sqrt(6)) / 4

What about a base fraction, like \u03c0/12? Knowing the exact value of that could help us find the exact value of a lot more angles. Here, we can use the difference formula.

What subtracts to equal \u03c0/12? \u03c0/3 and \u03c0/4 do!

sin(\u03b1 - \u03b2) = sin(\u03b1) * cos(\u03b2) - sin(\u03b2) * cos(\u03b1)
sin(\u03c0/12) = sin(\u03c0/3) * cos(\u03c0/4) - sin(\u03c0/3) * cos(\u03c0/4)
sin(\u03c0/12) = (sqrt(6) - sqrt(2)) / 4

Certainly isn\'t as nice looking as something like sin(\u03c0/3), which is why these values usually do not need to be memorized. This all goes to show that any trigonometric function will yield an exact value if the angle is in terms of \u03c0, due to the transcendentality of the trigonometric functions and the transcendentality of \u03c0.

sin(5 radians) will never be in exact value (it would be transcendental), but sin(3\u03c0/19) does have an exact value.

"},{"text":"

Simplification of the Sum and Difference Identities

Oftentimes, we find that the sum and difference identities are a bit too general for what we need to deal with. These can simplify to what are known as the double angle identities.

sin(\u03b1 + \u03b2) = sin(\u03b1) * cos(\u03b2) + sin(\u03b2) * cos(\u03b1)

What if \u03b1 = \u03b2?

sin(2\u03b1) = sin(\u03b1) * cos(\u03b1) + sin(\u03b1) * cos(\u03b1)
sin(2\u03b1) = 2sin(\u03b1) * cos(\u03b1)

Also note that cos(\u03b1 + \u03b2) = cos(\u03b1) * cos(\u03b2) - sin(\u03b1) * sin(\u03b2).

cos(2\u03b1) = cos(\u03b1) * cos(\u03b1) - sin(\u03b2) * sin(\u03b2)
cos(2\u03b1) = cos2(\u03b1) - sin2(\u03b1)

These can make it easier to expand out something like sin(3\u03b1).

sin(3\u03b1) = sin(2\u03b1 + \u03b1)
sin(2\u03b1 + \u03b1) = sin(2\u03b1) * cos(\u03b1) + sin(\u03b1) * cos(2\u03b1)
= 2sin(\u03b1) * cos(\u03b1) * cos(\u03b1) + sin(\u03b1) * (cos2(\u03b1) - sin2(\u03b1))
= 2sin(\u03b1) * cos2(\u03b1) + sin(\u03b1) * cos2(\u03b1) - sin(\u03b1) * sin2(\u03b1)
= 3sin(\u03b1) * cos2(\u03b1) - sin3(\u03b1)
= 3sin(\u03b1) * cos2(\u03b1) - sin3(\u03b1)
= 3sin(\u03b1) * (1-sin^2(\u03b1)) - sin3(\u03b1)
= 3sin(\u03b1) - 3sin3(\u03b1) - sin3(\u03b1)
sin(3\u03b1) = 3sin(\u03b1) - 4sin3(\u03b1)

Notice that there may be a relationship between the overall degree of the relationship and the inner coefficient of sin(3\u03b1). We will dive deeper into this in later articles.

"}]},"using-identities-single-sine-expression":{"title":"Using Identities: Single Sine Expression","sections":[{"text":"

Using Identities: Single Sine Expression

Alexander DeCarlo

As we journey through mathematics, we may find it very useful to rewrite trigonometric expressions as sine functions. This becomes helpful when solving trigonometric equations in some cases, but also with calculus and graphing. Let\'s see what we can do with a general sine expression.

The general form is A * sin(Bx + C) where A, B and C are constants. Let\'s try to separate this out. We want to work backwards from the answer because we are trying to convert complex expressions to this general form.

Using the sine addition identity, we can expand to A * [sin(Bx) * cos(C) + cos(Bx) * sin(C)]. Notice that sin(C) and cos(C) are constants. Carrying on, let\'s distribute out constant A and move the constants in front.

A * cos(C)sin(Bx) + A * sin(C)cos(Bx)

Since A * cos(C) and A * sin(B) are constant values, let\'s rewrite them as M and N respectively.

M * sin(Bx) + N * cos(Bx)
M = A * cos(C)
N = A * sin(C)

Hold on a second. We can use one of our identities to find a relationship between M, N, and A to remove those nasty sine and cosine functions.

M2 + N2 = (A * cos(C))2 + (A * sin(C))2
M2 + N2 = A2(cos2(C) + sin2(C))
M2 + N2 = A2

We have two more trivial relationships to uncover:

M = A * cos(C)
N = A * sin(C)

This means M / A = cos(C) and N / A = sin(C). Let\'s see what we managed to discover here.

M * sin(Bx) + N * cos(Bx) = A * sin(Bx + C)
M2 + N2 = A2
M / A = cos(C)
N / A = sin(C)

We can now turn any expression that involves adding cosine and sine with the same angle into a single sine function to be solved or graphed!

Let\'s look at the most simple of these expressions: cos(x) + sin(x).

M = 1
N = 1

(1)2 + (1)2 = A2

A = sqrt(2)
B = 1, since x * 1 = x in cos(x) and sin(x)

cos(C) = 1 / sqrt(2) = sqrt(2) / 2
C = \u03c0/4, 7\u03c0/4

sin(C) = sqrt(2) / 2
C = \u03c0/4, 3\u03c0/4

Here, we see that \u03c0/4 satisfies both of these relationships, so cos(x) + sin(x) = sqrt(2) * sin(x + \u03c0/4).

Let\'s try another!

4sqrt(3) * sin(5x) - 4cos(5x)

M = 4sqrt(3)
N = -4

16 + 48 = A2
A = 8

cos(C) = 4sqrt(3) / 8
sin(C) = -4/8

The value that satisfies both equations is 11\u03c0/6. Thus, 4sqrt(3) * sin(5x) - 4cos(5x) = 8sin(5x + 11\u03c0/6).

"}]},"solving-trigonometric-equations-ii":{"title":"Solving Trigonometric Equations II","sections":[{"text":"

Solving Trigonometric Equations II

Now that we have derived some identities and found some tricks within trigonometry, let\'s solve some equations that we could not previously solve before. Let\'s start with a problem involving sine and cosine.

2sin(4x) - 8cos(4x) = sin(2x) * cos(2x) + 3

This looks challenging, but it\'s quite simple with our knowledge of identities. First, we can use the single sine trick, then the product to sum identity. We could also multiply both sides of this equation by 2, and use the double angle identity. Both work here, but generally try to find the fastest solution for when things get harder.

4sin(4x) - 16cos(4x) = 2sin(2x) * cos(2x) + 6
4sin(4x) - 16cos(4x) = sin(4x) + 6
3sin(4x) - 16cos(4x) = 6

Now we can turn this into a single sine.

32 + 162 = A2

A = sqrt(265)
B = 4

3 / sqrt(265) = sin(C)
16 / sqrt(265) = cos(C)

C = 0.185 rad, where both equations are true.

sqrt(265) * sin(4x + 0.185) = 6
sin(4x + .185) = 6 / sqrt(265)

Remember, treat everything within sin() as u.

sin(u) = 6 / sqrt(265)

u = 0.377 + 2\u03c0n, where n is an integer.
u = 2.764 + 2\u03c0n

Substitute in for u.

4x + 0.185 = 0.377 + 2\u03c0n
4x = 0.192 + 2\u03c0n
x = 0.048 + \u03c0n / 2

4x + 0.185 = 2.764 + 2\u03c0n
x = 0.720 + \u03c0n / 2

Let\'s solve another, simpler equation for some exact values.

sin(x) * sin(2x) + cos(x) * cos(2x) = sqrt(3) / 2

Here, use the difference formula for cosine. When solving for equations, we should condense without losing a possible root.

cos(x - 2x) = sqrt(3) / 2
cos(-x) = sqrt(3) / 2

Remember that cosine isn\'t affected by that negative value, you can memorize this for efficiency, but it never hurts to sketch the unit circle to see this simple relationship.

cos(x) = sqrt(3) / 2

Here x = \u03c0/6, 11\u03c0/6.

Let\'s solve one more.

sin(x) + sin(3x) = cos(x) [0, 2\u03c0]

We can\'t use previous methods in this article, but we luck out because there isn\'t a constant term. This means if we can get a product form and factor, we can solve it. Good thing there\'s an identity for that!

2sin((x + 3x) / 2) * cos((x - 3x) / 2) = cos(x)
2sin(2x) * cos(-x) = cos(x)
2sin(2x) * cos(x) = cos(x)
2sin(2x) * cos(x) - cos(x) = 0
cos(x) * [2sin(2x) - 1] = 0
cos(x) = 0

x = \u03c0/2, 3\u03c0/2

sin(2x) = 1/2
2x = \u03c0/6, 5\u03c0/6

x = \u03c0/12, 5\u03c0/12 (add \u03c0 to navigate through the interval)
x = 13\u03c0/12, 17\u03c0/12

In the end, x = \u03c0/2, 3\u03c0/2, \u03c0/12, 5\u03c0/12, 13\u03c0/12, 17\u03c0/12.

"}]}}')}}]); \ No newline at end of file diff --git a/static/js/775.6bc52b72.chunk.js b/static/js/775.6bc52b72.chunk.js deleted file mode 100644 index d78454e..0000000 --- a/static/js/775.6bc52b72.chunk.js +++ /dev/null @@ -1 +0,0 @@ -"use strict";(self.webpackChunkmathinfo=self.webpackChunkmathinfo||[]).push([[775],{775:function(s){s.exports=JSON.parse('{"Summary":{"text":"

Proofs

","image":"
\\"\\"

Proofs graphic.

"},"pythagorean-theorem-and-distance-formula-derivation":{"title":"The Pythagorean Theorem and The Distance Formula Derivation","sections":[{"text":"

The Pythagorean Theorem and The Distance Formula Derivation

Alexander DeCarlo

The Pythagorean Theorem is one of the most fundamental theorems in all of Mathematics, but little are familiar with the proof of such a formula. This theorem is used to find the length of any diagonal line which gives us a formula for how to calculate the distance between two points in space. But how was it proven?

Let\'s throw away any knowledge of the Pythagorean Theorem and simply try to answer this question: What is the length of this line on the coordinate plane?

\'A

Looking at the image, we can see that A = (-6, -3), and B = (7, 6). We can easily calculate distance when the points are on the same dimension, but how do we calculate distance when two dimensions are involved? Looking at A and B, we can see they are 13 units away horizontally and 12 units away vertically. Let\'s represent this as a triangle.

","image":"
\\"\\"

Diagram of right triangle.

"},{"text":"

Here, c is our unknown, b is 13 and a is 12. This creates a right triangle. Right triangles are special for many reasons - a big factor being their properties of similarity.

Above shows how each component angle of each triangle must be all equivalent when Triangle 3 is bisected to where it internally creates Triangle 1 and Triangle 2. What similarity tells us is that the ratios of the side lengths must be equivalent. This is because while the triangles can have different sizes overall, their interior angles dictate the side length proportions. Therefore, when two angles are shown to be equal, the third must also be equal, creating these proportions. For example, in the previous model: a / b = e / d. Right triangles are the only triangles where you can have this and its two components similar, due to the creation of two equal degree angles when bisecting the hypotenuse. You can mess around with these ratios and find many relationships, so let\'s try to expand and discover something we didn\u2019t expect.

","image":"
\\"\\"

A triangle representation of the distance between points A and B.

"},{"text":"

Consider two ratios: a / c = e / a and b / c = f / b. These ratios are particularly interesting because they contain a and b side lengths of the main triangle twice in their respective equations. By cross multiplying, a2 = ce and b2 = cf. Since c is in both of these equations, let\'s add them together to simplify things: a2 + b2 = ce + cf. It is clear that we can factor out c: a2 + b2 = c(e + f). Looking at the diagram: e + f = c, meaning we can substitute e + f out for c and get a2 + b2 = c2. This relationship is not obvious at all, but can beautifully be derived from very simple similarity relationships. Some good news here is that since c is isolated, we can solve for the diagonal side length of a triangle given only vertical and horizontal lengths, being c = sqrt(a2 + b2). We can finally get that equation for the distance of any two points on a plane.



let D be the distance
C = sqrt(a2 + b2) (Substitute out hypotenuse length c.)
D = sqrt(a2 + b2) (a is our vertical length and b is our horizontal length.)
a = | y1 - y2 |
b = | x1 - x2 |
D = sqrt((y2 - y1)2 + (x2 - x1)2) (The absolute values are removed because the whole thing is squared, and since negative * negative = positive, the order of variables doesn\'t matter.)



Now we can find the distance between A = (-6, -3) and B = (7, 6). With proper substitution we get:



D = sqrt((-3 - 6)2 + (-6 - 7)2)
D = sqrt(81 + 169)
D = sqrt(250) = \u224815.811","image":""}]},"three-dimensional-pythagorean-theorem":{"title":"Three Dimensional Pythagorean Theorem","sections":[{"text":"

Three Dimensional Pythagorean Theorem

Alexander DeCarlo

Deriving the Pythagorean Theorem in three dimensions is surprisingly simple, as it builds on the 2D Pythagorean Theorem. The Pythagorean Theorem in two dimensions states: a2 + b2 = c2, where a and b are the legs of a right triangle, and c is the hypotenuse.

From this three dimensional illustration, we want to find what s is from non-angled lines: x, y and z.With help from the Pythagorean Theorem, we can deduce: c2 = x2 + y2. We can also notice that on the rectangular prism, z stays completely vertical. With that information, it is time to use the Pythagorean Theorem on the triangle containing s.



z2 + (sqrt(x2 + y2))2 = s2
x2 + y2 + z2 = s2 (Absolute value signs are not needed when canceling out due to x and y both squared.)



With changing the variable names to match the Pythagorean Theorem, we get: a2 + b2 + c2 = d2. There is a clear pattern when increasing one dimension, and it\'s by simply adding a new variable. To find a distance formula in 3D space, the process is exactly the same:



D = sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2)

When going up a dimension, all you need to do is add the subtracted coordinates squared. You can see this in action when going down dimensions. If D = sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2), decreasing to a single dimension gives us D = sqrt(x2 - x1)2), which when canceled is: D = | x2 - x1 |, the formula for finding the distance between two points on the number line.

Therefore, for n dimensions, the Pythagorean Theorem theoretically would be: a12 + \u2026 + an2 = c2, where n is the amount of dimensions, each term is the distance in its dimension, and c is the hypotenuse.

","image":"
\\"Diagram

Diagram of right triangle.

"}]}}')}}]); \ No newline at end of file diff --git a/static/js/775.bdfd4731.chunk.js b/static/js/775.bdfd4731.chunk.js new file mode 100644 index 0000000..6256141 --- /dev/null +++ b/static/js/775.bdfd4731.chunk.js @@ -0,0 +1 @@ +"use strict";(self.webpackChunkmathinfo=self.webpackChunkmathinfo||[]).push([[775],{775:function(s){s.exports=JSON.parse('{"Summary":{"text":"

Proofs

","image":"
\\"\\"

Proofs graphic.

"},"pythagorean-theorem-and-distance-formula-derivation":{"title":"The Pythagorean Theorem and The Distance Formula Derivation","sections":[{"text":"

The Pythagorean Theorem and The Distance Formula Derivation

Alexander DeCarlo

The Pythagorean Theorem is one of the most fundamental theorems in all of Mathematics, but little are familiar with the proof of such a formula. This theorem is used to find the length of any diagonal line which gives us a formula for how to calculate the distance between two points in space. But how was it proven?

Let\'s throw away any knowledge of the Pythagorean Theorem and simply try to answer this question: What is the length of this line on the coordinate plane?

\'A

Looking at the image, we can see that A = (-6, -3), and B = (7, 6). We can easily calculate distance when the points are on the same dimension, but how do we calculate distance when two dimensions are involved? Looking at A and B, we can see they are 13 units away horizontally and 12 units away vertically. Let\'s represent this as a triangle.

","image":"
\\"\\"

Diagram of right triangle.

"},{"text":"

Here, c is our unknown, b is 13 and a is 12. This creates a right triangle. Right triangles are special for many reasons - a big factor being their properties of similarity.

\'A

Above shows how each component angle of each triangle must be all equivalent when Triangle 3 is bisected to where it internally creates Triangle 1 and Triangle 2. What similarity tells us is that the ratios of the side lengths must be equivalent. This is because while the triangles can have different sizes overall, their interior angles dictate the side length proportions. Therefore, when two angles are shown to be equal, the third must also be equal, creating these proportions. For example, in the previous model: a / b = e / d. Right triangles are the only triangles where you can have this and its two components similar, due to the creation of two equal degree angles when bisecting the hypotenuse. You can mess around with these ratios and find many relationships, so let\'s try to expand and discover something we didn\'t expect.

","image":"
\\"\\"

A triangle representation of the distance between points A and B.

"},{"text":"

Consider two ratios: a / c = e / a and b / c = f / b. These ratios are particularly interesting because they contain a and b side lengths of the main triangle twice in their respective equations. By cross multiplying, a2 = ce and b2 = cf. Since c is in both of these equations, let\'s add them together to simplify things: a2 + b2 = ce + cf. It is clear that we can factor out c: a2 + b2 = c(e + f). Looking at the diagram: e + f = c, meaning we can substitute e + f out for c and get a2 + b2 = c2. This relationship is not obvious at all, but can beautifully be derived from very simple similarity relationships. Some good news here is that since c is isolated, we can solve for the diagonal side length of a triangle given only vertical and horizontal lengths, being c = sqrt(a2 + b2). We can finally get that equation for the distance of any two points on a plane.



let D be the distance
C = sqrt(a2 + b2) (Substitute out hypotenuse length c.)
D = sqrt(a2 + b2) (a is our vertical length and b is our horizontal length.)
a = | y1 - y2 |
b = | x1 - x2 |
D = sqrt((y2 - y1)2 + (x2 - x1)2) (The absolute values are removed because the whole thing is squared, and since negative * negative = positive, the order of variables doesn\'t matter.)



Now we can find the distance between A = (-6, -3) and B = (7, 6). With proper substitution we get:



D = sqrt((-3 - 6)2 + (-6 - 7)2)
D = sqrt(81 + 169)
D = sqrt(250) = \u224815.811","image":""}]},"three-dimensional-pythagorean-theorem":{"title":"Three Dimensional Pythagorean Theorem","sections":[{"text":"

Three Dimensional Pythagorean Theorem

Alexander DeCarlo

Deriving the Pythagorean Theorem in three dimensions is surprisingly simple, as it builds on the 2D Pythagorean Theorem. The Pythagorean Theorem in two dimensions states: a2 + b2 = c2, where a and b are the legs of a right triangle, and c is the hypotenuse.

\'A

From this three dimensional illustration, we want to find what s is from non-angled lines: x, y and z.With help from the Pythagorean Theorem, we can deduce: c2 = x2 + y2. We can also notice that on the rectangular prism, z stays completely vertical. With that information, it is time to use the Pythagorean Theorem on the triangle containing s.



z2 + (sqrt(x2 + y2))2 = s2
x2 + y2 + z2 = s2 (Absolute value signs are not needed when canceling out due to x and y both squared.)



With changing the variable names to match the Pythagorean Theorem, we get: a2 + b2 + c2 = d2. There is a clear pattern when increasing one dimension, and it\'s by simply adding a new variable. To find a distance formula in 3D space, the process is exactly the same:



D = sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2)

When going up a dimension, all you need to do is add the subtracted coordinates squared. You can see this in action when going down dimensions. If D = sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2), decreasing to a single dimension gives us D = sqrt(x2 - x1)2), which when canceled is: D = | x2 - x1 |, the formula for finding the distance between two points on the number line.

Therefore, for n dimensions, the Pythagorean Theorem theoretically would be: a12 + \u2026 + an2 = c2, where n is the amount of dimensions, each term is the distance in its dimension, and c is the hypotenuse.

","image":"
\\"Diagram

Diagram of right triangle.

"}]}}')}}]); \ No newline at end of file diff --git a/static/js/main.f168908d.js b/static/js/main.ad551be0.js similarity index 58% rename from static/js/main.f168908d.js rename to static/js/main.ad551be0.js index b8c59b8..156e1e2 100644 --- a/static/js/main.f168908d.js +++ b/static/js/main.ad551be0.js @@ -1,3 +1,3 @@ -/*! For license information please see main.f168908d.js.LICENSE.txt */ -!function(){var e={463:function(e,t,n){"use strict";var r=n(791),a=n(296);function o(e){for(var t="https://reactjs.org/docs/error-decoder.html?invariant="+e,n=1;n