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Trigonomotry – Study Guide

Degrees and Radians

You are probably familiar with talking about the measure of angles using degrees. Radians are another unit of measure for angles. One radian is the angle swept out by a circular arc that is the same length as its radius. A circle spans $360^{\circ}$ or $2\pi$ radians. An angle in degrees is typically represented as theta $(\theta)$ . An angle in radians is normally represented as x.

The conversion from degrees to radians is: $\dfrac{\theta}{360^{\circ}}=\dfrac{x}{2\pi}$ which reduces to $\dfrac{0}{180^{\circ}}=\dfrac{x}{\pi}$

To understand the representation of degrees to radians, a unit circle is represented below:

Exercises:
Convert to radians:	Convert to degrees:
$30^{\circ}=$	$\pi =$
$45^{\circ}=$	$\dfrac{\pi}{12} =$
$60^{\circ}=$	$\dfrac{\pi}{2} =$
	$\dfrac{3 \pi}{4} =$

Solutions:

$\dfrac{30^{\circ}}{180^{\circ}}=\dfrac{1}{6}=\dfrac{x}{\pi}, \quad x=\dfrac{\pi}{6}$
$\dfrac{45^{\circ}}{180^{\circ}}=\dfrac{1}{4}=\drrac{x}{\pi}, \quad x=\dfrac{\pi}{4}$
$\dfrac{60^{\circ}}{180^{\circ}}=\dfrac{1}{3}=\dfrac{x}{\pi}, \quad x=\dfrac{\pi}{3}$
$\dfrac{\pi}{\pi}=1=\dfrac{\theta}{180^{\circ}}, \quad \theta=\dfrac{180^{\circ}}$
$\dfrac{(\dfrac{\pi}{12})}{\pi}=\dfrac{1}{12}=\dfrac{\theta}{180^{\circ}}, \quad \theta = \dfrac{180^{\circ}}{12}=15^{\circ}$
$\dfrac{(\dfrac{\pi}{2})}{\pi}=\dfrac{1}{2}=\dfrac{\theta}{180^{\circ}}, \quad \theta = \dfrac{180^{\circ}}{2}=90^{\circ}$
$\dfrac{(\dfrac{3 \pi}{4})}{\pi}=\dfrac{3}{4}=\dfrac{\theta}{180^{\circ}}, \quad \theta=\dfrac{3\cdot 180^{\circ}}{4}=135^{\circ}$

The Unit Circle

The Unit Circle is a circle of radius 1 centered at the origin. Every point on the unit circle corresponds to an angle ?, formed by a line drawn from the origin to the point (?, ?), and the positive x-axis. Notice that drawing a vertical line down to the x-axis from the point (x,y) on the unit circle forms a right triangle with leg lengths x and y, and hypotenuse r =1.

Trigonometric Functions:

Angles in the Unit Circle:

Now we need a way to relate the coordinates of the coordinates of the point to the angle of the triangle made. We will use the functions sine, cosine, and tangent as shown below.

$\sin(\theta)=\frac{y}{r}$
$\cos(\theta)=\frac{x}{r}$
$\tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)}=\dfrac{\dfrac{y}{r}}{\dfrac{x}{r}}=\dfrac{y}{x}$

Another way to think of these trigonometric functions is to name the sides based on their location from the angle ? in the right triangle (see image above). They become:

$\mathrm{\mathbf{\underline{S}}in(\theta)=\mathbf{\underline{O}}pposite/\mathbf{\underline{H}}ypotenuse}$
$\mathrm{\mathbf{\underline{C}}os(\theta)=\mathbf{\underline{A}}djacent/\mathbf{\underline{H}}ypotenuse}$
$\mathrm{\mathbf{\underline{T}}an(\theta)=\mathbf{\underline{O}}pposite/\mathbf{\underline{A}}djacent}$

It may be helpful to remember this by memorizing the acronym SOH-CAH-TOA. These identities can be used to find sines and cosines of angles that are part of a right triangle.
Exercises
Use right-triangle trigonometry (SOH-CAH-TOA) to answer the following.
Problem 1:

Choose the correct response.

$\sin(\theta)=\frac{12}{13} \quad \cos(\theta)=\frac{5}{13} \quad \tan(\theta)=\frac{12}{5}$
$\sin(\theta)=\frac{5}{13} \quad \cos(\theta) = \frac{12}{13} \quad \tan(\theta)=\frac{5}{12}$
$\sin(\theta)=\frac{5}{13} \quad \cos(\theta) = \frac{12}{13} \quad \tan(\theta)=\frac{12}{5}$
$\sin(\theta)=\frac{12}{13} \quad \cos(\theta) = \frac{5}{13} \quad \tan(\theta)=\frac{5}{12}$

Problem 2:

Choose the correct response.

$\sin(\theta)=\frac{8}{17} \quad \cos(\theta)=\frac{15}{17} \quad \tan(\theta)=\frac{15}{8}$
$\sin(\theta)=\frac{15}{17} \quad \cos(\theta) = \frac{8}{17} \quad \tan(\theta)=\frac{15}{8}$
$\sin(\theta)=\frac{15}{17} \quad \cos(\theta) = \frac{8}{17} \quad \tan(\theta)=\frac{8}{15}$
$\sin(\theta)=\frac{8}{17} \quad \cos(\theta) = \frac{15}{17} \quad \tan(\theta)=\frac{8}{15}$

Problem 3:

Choose the correct response.

$\sin(\theta)=\frac{24}{25} \quad \cos(\theta)=\frac{7}{25} \quad \tan(\theta)=\frac{7}{24}$
$\sin(\theta)=\frac{24}{25} \quad \cos(\theta) = \frac{7}{25} \quad \tan(\theta)=\frac{24}{7}$
$\sin(\theta)=\frac{7}{25} \quad \cos(\theta) = \frac{24}{25} \quad \tan(\theta)=\frac{7}{24}$
$\sin(\theta)=\frac{7}{25} \quad \cos(\theta) = \frac{24}{25} \quad \tan(\theta)=\frac{24}{7}$

Solutions:
Problem 1: A
Problem 2: D
Problem 3: B
Special Angles:

The angles that lie along an axis are special because you cannot make a picture of a triangle for them to find the values. Fortunately the coordinates are easier to visualize and still apply.

$\cos(0^{\circ})=\frac{x}{r}=\dfrac{x-axis,length}{radius}=\frac{1}{1}=1$

$\sin(0^{\circ})=\frac{y}{r}=\dfrac{y-axis,length}{radius}=\frac{0}{1}=0$

The thing to watch out for is division by zero, such as this example:
$\tan(90^{\circ})=\frac{y}{x}=\dfrac{y-axis,length}{x-axis,length}=\frac{1}{0}=undefined$
Quadrants:

The circle is divided up into 4 regions called quadrants. In quadrant I both x and y are positive. In quadrant II x is negative and y is positive. In quadrant III both x and y are negative. In quadrant IV x is positive and y is negative. Since sin, cos, and tan involve x and y coordinates these coordinates retain their positive or negative sign. Note that the radius, r, is always positive.

$\sin(\theta)=\frac{x}{r}$ (positive)
$\cos(\theta)=\frac{x}{r}$ (positive)
$\tan(\theta)=\frac{y}{x}$ (positive)

Quadrant II (-, +):

$\sin(\theta)=\frac{y}{r}$ (positive)
$\cos(\theta)=\frac{x}{r}$ (negative)
$\tan(\theta)=\frac{y}{x}$ (negative)

Quadrant III (-,-):

$\sin(\theta)=\frac{y}{r}$ (negative)
$\cos(\theta)=\frac{x}{r}$ (negative)
$\tan(\theta)=\frac{y}{x}$ (positive)

Quadrant IV (+,-):

$\sin(\theta)=\frac{y}{r}$ (negative)
$\cos(\theta)=\frac{x}{r}$ (positive)
$\tan(\theta)=\frac{y}{x}$ (negative)

Another way to approach the topic of sign is to just look at each function individually.

$\sin(\theta)=\frac{y}{r}$ is positive in the I and II quadrants and negative in the III and IV quadrants.
$\cos(\theta)=\frac{x}{r}$ is positive in the I and IV quadrants and negative in the II and III quadrants.
$\tan(\theta)=\frac{y}{x}$ is positive in the I and III quadrants and negative in the II and IV quadrants.

Exercises:

Find $\sin(\theta), \cos(\theta),\mathrm{and} \tan(\theta)$ .

Solution:

In the second quadrant,
$\sin(\theta)=\frac{y}{r}=\frac{4}{5}$ (positive) $\cos(\theta)=\frac{x}{r}=\frac{3}{5}$ (negative) $\tan(\theta)=\frac{y}{x}=\frac{4}{3}$ (negative)
Reference Angles:

Every angle over $90^{\circ}$ has a corresponding reference angle in the first quadrant (0, -90°). A reference angle is an angle that has the same numerical values for its trigonometric functions but not necessarily the same sign (+,-). There are a few different ways to find reference angles. Numerically you can just memorize a rule for every quadrant:

Quadrant I: $\theta$
Quadrant II: $180-\theta$
Quadrant III: $\theta-180$
Quadrant IV: $360-\theta$

Graphically you can find the reference angle by reflecting over the axis to get back to the first quadrant.

The angle in quadrant II can be flipped over the y-axis into quadrant I.
The angle in quadrant IV can be flipped over the x-axis into quadrant I.
The angle in quadrant III must be flipped twice, once over the x-axis and once over the y-axis in any order to get to quadrant I. Alternatively the line made from the angle in quadrant III can just be extended into quadrant I.

Exercises:
Fill out the table with the appropriate reference angle.

Angle	Reference Angle
$0^{\circ}$
$30^{\circ}$
$45^{\circ}$
$60^{\circ}$
$90^{\circ}$
$120^{\circ}$
$135^{\circ}$
$150^{\circ}$
$180^{\circ}$
$210^{\circ}$
$225^{\circ}$
$240^{\circ}$
$270^{\circ}$
$300^{\circ}$
$315^{\circ}$
$330^{\circ}$
$360^{\circ}$

Solutions:

Angle	Reference Angle
$0^{\circ}$	$0^{\circ}$
$30^{\circ}$	$30^{\circ}$
$45^{\circ}$	$45^{\circ}$
$60^{\circ}$	$60^{\circ}$
$90^{\circ}$	$90^{\circ}$
$120^{\circ}$	$60^{\circ}$
$135^{\circ}$	$45^{\circ}$
$150^{\circ}$	$30^{\circ}$
$180^{\circ}$	$180^{\circ}$
$210^{\circ}$	$30^{\circ}$
$225^{\circ}$	$45^{\circ}$
$240^{\circ}$	$60^{\circ}$
$270^{\circ}$	$270^{\circ}$
$300^{\circ}$	$60^{\circ}$
$315^{\circ}$	$45^{\circ}$
$330^{\circ}$	$30^{\circ}$
$360^{\circ}$	$0^{\circ}$

Exercises:
Problem 1: Find the reference angle associated with $120^{\circ}$ and for $\cos(120^{\circ})$ .
Problem 2: Find the reference angle associated with $210^{\circ}$ and solve for $\sin(210^{\circ})$ .
Problem 3: Find the reference angle associated with $315^{\circ}$ and solve for $\tan(315^{\circ})$ .

Solutions

Problem 1:
Step 1: Draw a picture.

Step 2:

The angle is in quadrant II so reference angle is $180^{\circ}-\theta = 180^{\circ}-120^{\circ}=60^{\circ}$ . Recall CAH and as you can see in the picture above this would make $\cos(60^{\circ})=\frac{1}{2}$ . In quadrant II, cos is negative so $\cos(120^{\circ})=\frac{-1}{2}$ .

Problem 2:
Step 1: Draw a picture.

Step 2:

The angle is in quadrant III so the reference angle is $\theta -180^{\circ}=210^{\circ}-180^{\circ}=30^{\circ}$ . Recall SOH and as you can see in the picture above this would make $\sin(30^{\circ})=\frac{1}{2}$ . In quadrant III sin is negative so $\sin(210^{\circ})=\frac{-1}{2}$ .

Problem 3

Draw a picture.

Problem3-Step 2:

The angle is in quadrant IV so the reference is angle $360^{\circ}-\theta=360^{\circ}-316^{\circ}=45^{\circ}$ . Recall TOA and as you can see in the picture above this would make $/tan(45^{\circ})=1$ . In quadrant IV tan is negative so $\tan(315^{\circ})=-1$ .
Exercises
Use your knowledge of special angles and reference angles we covered earlier with the new material on special triangles and to fill out this table (Note: Some teachers may want you to be able to fill this chart out from memory on a quiz or test.)

Angle	sin θ	cos θ	tan θ
$0^{\circ}$
$30^{\circ}$
$45^{\circ}$
$60^{\circ}$
$90^{\circ}$
$120^{\circ}$
$135^{\circ}$
$150^{\circ}$
$180^{\circ}$
$210^{\circ}$
$225^{\circ}$
$240^{\circ}$
$270^{\circ}$
$300^{\circ}$
$315^{\circ}$
$330^{\circ}$
$360^{\circ}$

Solutions:

Angle	sin θ	cos θ	tan θ
$0^{\circ}$	0	1	0
$30^{\circ}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$
$45^{\circ}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	1
$60^{\circ}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
$90^{\circ}$	1	0	undefined
$120^{\circ}$	$\frac{\sqrt{3}}{2}$	$\frac{-1}{2}$	$-\sqrt{3}$
$135^{\circ}$	$\frac{\sqrt{2}}{2}$	$\frac{-\sqrt{2}}{2}$	-1
$150^{\circ}$	$\frac{1}{2}$	$\frac{-\sqrt{3}}{2}$	$\frac{-\sqrt{3}}{3}$
$180^{\circ}$	0	-1	0
$210^{\circ}$	$\frac{-1}{2}$	$\frac{-\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$
$225^{\circ}$	$\frac{-\sqrt{2}}{2}$	$\frac{-\sqrt{2}}{2}$	1
$240^{\circ}$	$\frac{-\sqrt{3}}{2}$	$\frac{-1}{2}$	$\sqrt{3}$
$270^{\circ}$	-1	0	Undefined
$300^{\circ}$	$\frac{-\sqrt{3}}{2}$	$\frac{1}{2}$	$-\sqrt{3}$
$315^{\circ}$	$\frac{-\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	-1
$330^{\circ}$	$\frac{-1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{-\sqrt{3}}{3}$
$360^{\circ}$	0	1	0

Reciprocal Trigonometric functions

In addition to the three main trigonometric functions there are three reciprocal trigonometric functions. They are the secant, cosecant, and cotangent functions. They have the same sign rules as their normal trigonometric functions based on the quadrant the angle is in. THey also have the same reference angle rules.

$\csc(\theta)=\dfrac{1}{\sin(\theta)}=\dfrac{r}{y}$
$\sec(\theta)=\dfrac{1}{\cos(\theta)}=\dfrac{r}{x}$
$\cot(\theta)=\dfrac{1}{\tan(\theta)}=\dfrac{\cos(\theta)}{\sin(\theta)}=\dfrac{x}{y}$

Exercises:
The reciprocal trigonometric functions are not very different from the original three and so we can make another table and find all of the values for these functions.

Angle	csc θ	sec θ	cot θ
$0^{\circ}$
$30^{\circ}$
$45^{\circ}$
$60^{\circ}$
$90^{\circ}$
$120^{\circ}$
$135^{\circ}$
$150^{\circ}$
$180^{\circ}$
$210^{\circ}$
$225^{\circ}$
$240^{\circ}$
$270^{\circ}$
$300^{\circ}$
$315^{\circ}$
$330^{\circ}$
$360^{\circ}$

Solutions

Angle	csc θ	sec θ	cot θ
$0^{\circ}$	Undefined	1	Undefined
$30^{\circ}$	2	$\frac{2\sqrt{3}}{3}$	$\sqrt{3}$
$45^{\circ}$	$\sqrt{2}$	$\sqrt{2}$	1
$60^{\circ}$	$2\sqrt{3}}{3}$	2	$\frac{\sqrt{3}}{3}$
$90^{\circ}$	1	Undefined	0
$120^{\circ}$	$\frac{2\sqrt{3}}{3}$	-2	$\frac{-\sqrt{3}}{3}$
$135^{\circ}$	$\sqrt{2}$	$-\sqrt{2}$	-1
$150^{\circ}$	2	$\frac{-2\sqrt{3}}{3}$	$-\sqrt{3}$
$180^{\circ}$	Undefined	-1	Undefined
$210^{\circ}$	-2	$\frac{-2\sqrt{3}}{3}$	$\sqrt{3}$
$225^{\circ}$	$-\sqrt{2}$	$-\sqrt{2}$	1
$240^{\circ}$	$\frac{-2\sqrt{3}}{3}$	-2	$\frac{\sqrt{3}}{3}$
$270^{\circ}$	-1	Undefined	0
$300^{\circ}$	$\frac{-2\sqrt{3}}{3}$	2	$\frac{-\sqrt{3}}{3}$
$315^{\circ}$	$-\sqrt{2}$	$\sqrt{2}$	-1
$330^{\circ}$	-2	$\frac{2\sqrt{3}}{3}$	$-\sqrt{3}$
$360^{\circ}$	Undefined	1	Undefined

Graphs of Trigonometric Functions

Let’s take a moment to graph sin, cos, tan, csc, sec, and cot as functions of angle. Put your graphing calculator in radian mode, and use windows y:[-2,2] x:[-5,5].
Solutions:

You may have noticed that these functions repeated themselves, and some of them look very similar. Sine and cosine graphs are actually the same graph, shifted out of phase by $\frac{\pi}{2}$ . The graphs are periodic, meaning that they repeat every $2\pi$ radians. As would be expected, cosecant and secant have the same behavior. Note that sine and cosine have a range of [-1],1] while secant and cosecant have a range of $(-\infty,-1]\cup [1,\infty)$ and tangent and cotangent have a range from $(-\infty,\infty)$ but have a restricted domain.

(1) $\begin{equation*} \cos(\theta)=\sin(\theta + 90^{\circ})\end{equation*}$

(2) $\begin{equation*} \sin(\theata)=\cos(\theta - 90^{\circ})\end{equation*}$

(3) $\begin{equation*} \sec(\theta)=\csc(\theta + 90^{\circ})\end{equation*}$

(4) $\begin{equation*} \csc(\theta)=\sec(\theta - 90^{\circ}) \end{equation*}$

Graphical Representation of Trigonometric functions:

Notice that drawing a vertical line down to the x-axis from the point (x,y) on the unit circle forms a right triangle with leg lengths x and y, and hypotenuse r. The angle created is ?. Because the radius of the unit circle is 1, we can use right-angle trigonometry to determine that ? = cos(θ), and ? = sin(?). You should know how to represent sine and cosine on the
unit circle, as shown below.

Cosecant, secant, tangent and cotangent can also be represented on the unit circle. You will probably not need to know how to do this. If you are curious, see the figure below for an example. If you are not, feel free to skip ahead.

Inverse Trig Functions:

The inverse trigonometric functions are used to find the angle given a value
They are either represented as:

$\sin^{-1}(\frac{y}{r})=\theta$
$\cos^{-1}(\frac{x}{r})=\theta$
$\tan^{-1}(\frac{y}{x})=\theta$

Or to avoid confusion they can also be represented as:

$\arcsin(\frac{y}{r})=\theta$
$\arccos(\frac{x}{r})=\theta$
$\arctan(\frac{y}{x})=\theta$

There is one subtlety to using these functions. Because the trigonometric functions are periodic, there are multiple answers to the inverse functions. When you enter them into a calculator, it returns an answer between – ? and ? radians.

Equations to Solve Angles and Sides of Triangles

Law of sines

For triangles other than right triangles, right-angle trigonometry cannot be used to find the value of sines and cosines. For finding information about these triangles, the Law of Sines can be very useful. The Law of Sines states that the ratio of the sine of an angle to
the length of its opposite side is the same for all angles in a triangle. $\dfrac{\sin(A)}{a}=\dfrac{\sin(B)}{b}=\dfrac{\sin(C)}{c}$
Exercise:
Given:
C= $26^{\circ}$
c=3.2 cm
a=6.7 cm
Find A, B, and b
(Use the picture above)
Step 1: Set up the Line of Sines
$C=26^{\circ}$ ,

$\dfrac{\sin(26^{\circ})}{3.2}=\dfrac{\sin(A)}{6.7}$
Step 2: Rearrange to solve for C
$\arcsin(\dfrac{6.7\sin(26^{\circ})}{3.2})=66.6^{\circ}$
Note: $\arcsin({\theta})$ is positive in quadrants I and II so A is actually either $66.6^{\circ}$ or $180^{\circ}-66.6^{\circ}=113.4^{\circ}$ but we know that we are looking at an acute angle from the picture.

Step 3: solve for missing angle

$\mathrm{B}=180^{\circ}-66.6^{\circ}-26^{\circ}=87.4^{\circ}$
Step 4: Solve for remaining side
$\dfrac{\sin(26^{\circ})}{3.2}=\dfrac{\sin(87.4^{\circ})}{a}$

$a=\dfrac{3.2 \cdot \sin(87.4)}{\sin(26^{\circ})}=7.3 \mathrm{cm}$
Law of Cosines

If you are given a triangle with the measure of an angle and the two non-corresponding sides, the Law of Cosines will allow you to solve for the unknowns. Notice that once again, the side called a is opposite angle A, b is the side opposite angle B, and c is the side opposite angle C.

Note: If there is a right angle in the triangle, we call that angle C . This makes c the hypotenuse and since cos(C)=0, the Law of Cosines simplifies to the Pythagorean Theorem.

Exercise:
Given:
$\matnrm{A=71^{\circ}}$
$\mathrm{c=5.9 cm}$
$\mathrm{b=2.8 cm}$
Find a, B and C using the Law of Cosines
(Use picture above.)
Solution:
Step 1: Set up Equation

$5.9^2+2.8^2- \cdot 5.9 \cdot 2.8 \cdot \cos(71^{\circ})+a^2$
Step 2: Rearrange for a
$a = \pm \sqrt{5.9^2+2.8^2-2 \cdot 5.9 \cdot 2.8 \cdot \cos(71^{\circ})=5.65 \mathrm{cm}}$
Step 3: Set up equation for another angle
$2.8^2+5.65^2-2 \cdot 2.8 \cdot 5.65 \cdot \cos(B)=5.9^2$
Step 4: Rearrange for B
$B=\arccos(\dfrac{2.8^2+5.65^2-5.9^2}{2\cdot 2.8 \cdot 5.65})=81.4^{\circ}$
Step 5: Solve for the remaining anble
$C=180^{\circ}-81.4^{\circ}-71^{\circ}=27.6^{\circ}$
Heron’s Formula:

To find the area of a triangle you would normally use the formula $\mathrm{A=\frac{1}{2}=bh}$ . To use this formula, you must know or calculate the height of the triangle. Heron’s Formula allows us to calculate the area of the triangle using the lengths of all three sides, without finding the height.

$Area=\sqrt{|s(s-a)(s-b)(s-c)|}$
$s=\frac{1}{2}(a+b+c)$

Exercise:
Given the information in the Figure below, find the area of the triangle:
Step 1: Find s
$s=\frac{1}{2}(2.7+3.2+6.9)=6.4 \mathrm{cm}$
Step 2: Plug in values
$A=\sqrt{|6.4(6.4-2.7)(6.4-3.2)(6.4-6.9)|}=4.87 \mathrm{cm}^2$
Pythagorean Theorem:

We have talked about the Pythagorean Theorem and how it relates to a unit circle, now we are going to look at how it relates to sine and cosine. The Pythagorean Identity is just as important as the theory and is one you should memorize. It is derived from the following equations:

$a^2+b^2=c^2$
$x^2+y^2=1$
$\cos(\theta)=x$
$\sin(\theta)=y$
So if we substitute into the second equation we get:
$\sin^2(\theta)+\cos^2(\theta)=1$
Identities Derived from Pythagorean Theorem:

The Pythagorean identity, sin2 ? + cos2 ? = 1, can be used to obtain more useful identities. If you have a problem that involves a tangent or cotangent squared term these two identities will be useful:

Step 1: Start with the Pythagorean Identity. Divide by $\sin^2(\theta)$ .

$\frac{(\sin^2(\theta)+\cos^2(\theta)+1)}{\sin^2(\theta)}$
Step 2: Rewrite the Problem.
$\dfrac{\sin^2(\theta)}{\sin^2(\theta)}+\dfrac{\cos^2(\theta)}{sin^2(\theta)}=\dfrac{1}{\sin^2(\theta)}$
Step 3: Reduce fractions into simplest terms.
$\boxed{\mathbf{1+\cot^2(\theta)=\csc^2(\theta)}}$
Step 1: Start with the Pythagorean Identity. Divide by $\cos^2(\theta)$
$\dfrac{\sin^2(\theta)+\cos^2(\theta)=1}{\cos^2(\theta)}$
Step 2: Rewrite the problem (in terms of simpler terms)
$\dfrac{\sin^2(\theta)}{\cos^2(\theta)}+\dfrac{\cos^2(\theta)}{\cos^2(\theta)}=\dfrac{1}{\cos^2(\theta)}$
Step 3: Reduce fractions to simplest terms
$\boxed{\mathbf{\tan^2(\theta)+1=\sec^2(\theta)}}$

Angle Identities

Negative Angle Identities

To find the negative angle identities, graphing the functions is helpful. First plot the normal function on both sides of the y axis and then reflect over the axis to see how they relate. If you have a graphing calculator you can just plot with a negative theta and compare to the normal graph. The graphs are provided for your review below along with the algebraic identities.

$\sin(-\theta)=-\sin(\theta)$
$\cos(-\theta)=-\cos(\theta)$
$\tan(-\theta)=-\tan(\theta)$

Sum and Difference Identities:

The sum and difference identities are derived from the unit circle using geometric principles. They are very useful for finding values of trigonometric functions that you normally would need a calculator for. You can use sines and cosines you already have memorized to calculate other values. For example, $\sin(15^{\circ})$ can be calculated as $\sin(45^{\circ}-30^{\circ})$ .

$\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)\\ \sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)\\ \cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)\\ \cos(A-B)=\cos(A)cos(B)+\sin(A)\sin(B) \tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} \tan(A-B)=\dfrac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)}$

Exercises:

Find the $\sin(15^{\circ})$ by using a difference equation and using values you already know.
Find the $\cos(75^{\circ})$ by using a difference equation and using values you already know.
Find the $\tan(105^{\circ})$ by using a sum equation and using values you already know.

Solutions:
Problem 1:
$\sin(15^{\circ})=\sin(45^{\circ}-30^{\circ})=\sin(45^{\circ})\cos(30^{\circ})-\cos(45^{\circ})\sin(30^{\circ})=\dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6}}{4}-\dfrac{\sqrt{2}}{4}=0.25882$
Problem 2:
$\cos(75^{\circ})=\cos(135^{\circ}-60^{\circ})=\cos(135^{\circ})\cos(60^{\circ})+\sin(135^{\circ})\sin(60^{\circ})=-\dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2}=-\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{6}}{4}=0.25882$
Problem 3:
$\tan(105^{\circ})=\tan(60^{\circ}+45^{\circ})=\dfrac{\tan(60^{\circ})+\tan(45^{\circ})}{1-\tan(60^{\circ})\cdot \tan(45^{\circ})} = \dfrac{\sqrt{3}+1}{1-\sqrt{3} \cdot 1}=-3.73$
Product to Sum Identities
Converting from a product to a sum can be a useful shortcut for simplifying expressions.

$\setlength{\parindent}{0pt} \noIndent \sin(A)\sin(B)=\frac{1}{2}[cos(A-B)-cos(A+B)]\\ \cos(A)\cos(B)=\frac{1}{2}[\cos(A-B)+\cos(A+B)]\\ \cos(A)\sin(B)=\frac{1}{2}[\sin(A+B)-\sin(A-B}] \\ \sin(A)\cos(B)=\frac{1}{2}[\sin(A+B)+\sin(A-B)]$

Exercise:
$\sin(45^{\circ})\cos(15^{\circ})=$
Solution:
$\sin(45^{\circ})\cos(15^{\circ})=\frac{1}{2}[\sin(45^{\circ}+15^{\circ})+\sin(45^{\circ}-15^{\circ})]=\frac{1}{2}[\frac{\sqrt{3}}{2}+\frac{1}{2}]=\frac{\sqrt{3}}{4}+\frac{1}{4}=0.683$

Sum to Product Identities

Now we need a way to go from a sum to a product. We will use the product to sum identities and work backwards to get these new sums to product identities. The first thing we need to fix is that our side with the sum has A-B inside the function rather than just an A or B so we will need to substitute in to get a single variable.

Let $A=\dfrac{A+B}{2}$ and let $B=\dfrac{A-B}{2}$ and substituting into the following equations:
Start with a product sum identity:

$\sin(A)\cos(B)=\frac{1}{2}[\san(A+B)+\sin(A-B)]$
Use Substitution:
$\sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})=\dfrac{1}{2}[\sin(\dfrac{A+B}{2}+\dfrac{A-B}{2})+\sin(\dfrac{A+B}{2}-\dfrac{A-B}{2})]$
Simplify:
$\sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})=\frac{1}{2}[\sin(A)+\sin(B)]$
Start with a product sum identity:
$\cos(A)\cos(B)=\frac{1}{2}[\cos(A-B)+\cos(A+B)]$
Use substitution:
$\cos(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})=\frac{1}{2}[\cos(\dfrac{A+B}{2}-\dfrac{A-B}{2})+\cos(\dfrac{A+B}{2}+\dfrac{A-B}{2})]$
Simplify:
$\cos(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})=\frac{1}{2}[\cos(A)+\cos(B)]$
Starting with a product sum identity:
$\sin(A)\sin(B)=\frac{1}{2}[\cos(A-B)+\cos(A+B)]$
Use substitution:
$\sin(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2})=\frac{1}{2}[\cos(\dfrac{A+B}{2}-\dfrac{A-B}{2})+\cos(\dfrac{A+B}{2}+\dfrac{A-B}{2})]$
Simplify:
$\sin(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2})=\frac{1}{2}[\cos(A)+\cos(B)]$
Starting with product sum identity:
$\cos(A)\sin(B)=\frac{1}{2}[\sin(A+B)-\sin(A-B)]$
Use substitution:
$\cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2})=\frac{1}{2}[\sin(\dfrac{A+B}{2}+\dfrac{A-B}{2})-\sin(\dfrac{A+B}{2}-\dfrac{A-B}{2})]$
Simplify:
$\cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2})=\frac{1}{2}[\sin(A)-\sin(B)]$

Multiplying these last equations by 2 on each side to come up with the final product identities gives us:

$\sin(A)+\sin(B)=2\sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})\\ \cos(A)+\cos(B)=2\cos(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})\\ \cos(A)-\cos(B)=-2\sin(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2})\\ \sin(A)-\sin(B)=2\cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2})$

Double Angle Identities:
The double angle identities are derived using the sum and difference identities as shown below.

$\sin(2A)=\sin(A+A)=\sin(A)\cos(A)+\cos(A)\sin(A)=2\sin(A)\cos(A)\\ \cos(2A)=\cos(A+A)=\cos(A)\cos(A)-\sin(A)\sin(A)=\cos^2(A)-\sin^2(A)\\ \tan(2A)=\tan(A+A)=\dfrac{\tan(A)+\tan(A)}{1-\tan(A)\tan(A)}=\dfrac{2\tan(A)}{1-\tan^2(A)}$
These equations SIMPLIFY to become:

$\sin(2A)=2\sin(A)\cos(A)\\ \cos(2A)=1-2\sin^2(A)=2\cos^2(A)-1=\cos^2(A)-\sin^2(A)\\ \tan(2A)=\dfrac{2\tan(A)}{1-\tan^2(A)}$

Half Angle Identities:

You may have noticed that in the double angle identities section, the cosine identity had a couple of other alternate forms that were not steps in the derivation. We did this because they are used in the derivation of the half angle formulas as shown below. Note: we have replaced A with x to avoid confusion in later steps.

$\cos(2x)=1-2\sin^2(x)\\ \sin^2(x)=\dfrac{1-\cos(2x)}{2}\\ \sin(x)=\pm \sqrt{\dfrac{1-\cos(2x)}{2}$

Let $x=\frac{A}{2}$ and we now have an equation for $\sin(\frac{A}{2})$ in terms of functions of just A.

$\cos(2x)=2\cos^2(X)-1\\ \cos(x)=\pm \sqrt{\dfrac{1+\cos(2x)}{2}}$

Finishing the substitution with 2x=A, this gives us the final equations:

$\sin(\dfrac{A}{2})=\pm \sqrt{\dfrac{1-\cos(A)}{2}}\\ \cos(\frac{A}{2})=\pm \sqrt{\dfrac{1+\cos(A)}{2}}$

To get a half angle identity for tangent we use the previous two identities we have already found.

$\tan(\frac{A}{2})=\dfrac{\sin(\frac{A}{2})}{\cos(\frac{A}{2})}=\dfrac{\pm \sqrt{\dfrac{1-\cos(A)}{2}}}{\pm \sqrt{\dfrac{1+\cos(A)}{2}}}=\pm \sqrt{\dfrac{1-\cos(A)}{1+\cos(A)}}$
$\tan(\frac{A}{2})=\dfrac{\sin(\frac{A}{2})}{\cos(\frac{A}{2})}\cdot \dfrac{2\cos(\frac{A}{2})}{2\cos(\frac{A}{2})}=\dfrac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{2\cos^2(\frac{A}{2})}=\dfrac{\sin(2\cdot \frac{A}{2})}{1+\cos(2 \cdot{A}{2})}=\dfrac{\sin(A)}{1+\cos(A)}$
$\tan(\frac{A}{2})=\dfrac{\sin(\frac{A}{2})}{\cos(\frac{A}{2})}\cdot \dfrac{2\sin(\frac{A}{2})}{2\sin(\frac{A}{2})}=\dfrac{2\sin^2(\frac{A}{2})}{2\sin(\frac{A}{2})\cos(\frac{A}{2})}=\dfrac{1-\cos(2\cdot \frac{A}{2})}{\sin(2\cdot \frac{A}{2})}=\dfrac{1-\cos(A)}{\sin(A)}$

Other Identities

These are more obscure identities that you may need at some point for reference and we will list them so that you know where you can find find them and can have an idea of what is out there.

Square Identities
$\sin^2(A)=\dfrac{1-\cos(2A)}{2}$
$\cos^2(A)=\dfrac{1+\cos(2A)}{2}$
$\tan^2(A)=\dfrac{1-\cos(2A)}{1+\cos(2A)}$

Inverse Identities:
$\arcsin(x)+\arccos(x)=\frac{\pi}{2}$
$\arctan(x)+\arccot(x)=\frac{\pi}{2}$
$\arctan(x)+\arctan(\frac{1}{x}=\frac{\pi}{2}, \mathrm{if\quad x>0}$
$\arctan(x)+\arctan(\frac{1}{X})=-\frac{\pi}{2}, \mathrm{if\quad x<0}$

Verifying Identities:

Any true equation that puts the trigonometric functions into different forms is an identity. Some of these are more useful than others. Of the ones listed above, you will probably only need to know or know how to get the Pythagorean Identities. The others are normally given to you in an equation sheet and you just need to know how to use them. When practicing trigonometric identities you will be asked to verify that a given equation is an identity. To do this, you will need to work with one side of the equation only and end up with the other side of the equation. These can be fairly simple or extremely complex. We will work quite a few of these so that you can get as much experience as you would like working with identities.

Hint: Start working with the most complicated side, it normally will make it easier.

Verify the following identities.

$\tan(\theta)+1=\sec(\theta)(\sin(\theta)+\cos(\theta))$

Solution:

Step 1: Write everything in terms of sin and cos on one side.

$\tan(\theta)+1=\frac{1}{\cos(\theta)}(\sin(\theta)+\cos(\theta))$

Step 2: Rearrange equation.

$\tan(\theta)+1=\dfrac{\sin(\theta)}{\cos(\theta)}+\dfrac{\cos(\theta)}{\cos(\theta)}$

Step 3: Reduce if possible.

$\tan(\theta)+1=\tan(\theta)+1$

Step 4: If both sides of the equation match, it is an identity.

$\sec^2(\theta)(1-\cos^2(\theta))=\dfra{\sin^2(\theta)}{1-\sin^2(\theta)}$

Solution:

Step 1: Write everything in terms of sin and cos on one side.

$\frac{1}{\cos^2(\theta)}(1-\cos^2(\theta))=\dfrac{\sin^2(\theta)}{1-\sin^2(\theta)}$

Step 2: Rearrange equation.

$\dfrac{1}{\cos^2(\theta)}-\dfrac{\cos^2(\theta)}{\cos^2(\theta)}=\dfrac{\sin^2(\theta)}{1-\sin^2(\theta)}$

Step 3: Reduce if possible.

$\sec^2(\theta)-\dfrac{\sin^2(\theta)}{1-\sin^2(\theta)}$

Step 4: Apple known identities $\tan^2(\theta)+1=\sec^2(\theta)$ and rearrange to suite equation $\sec^2(\theta)-1=\tan^2(\theta)$

$\tan^2(\theta)=\dfrac{\sin^2(\theta)}{1-\sin^2(\theta)}$
$\dfrac{\sin^2(\theta)}{\cos^2(\theta)}=\dfrac{\sin^2(\theta)}{1-\sin^2(\theta)}$
$\sin^2(\theta)+\cos^2(\theta)=1$
$\cos^2(\theta)=1-\sin^2(\theta)$
$\dfrac{\sin^2(\theta)}{1-\sin^2(\theta)}=\dfrac{\sin^2(\theta)}{1-\sin^2(\theta)}$

Step 5: If both sides of the equation match, it is an identity.

$\dfrac{\sin^2(\theta)-1}{\sin^2(\theta)\cot^2(\theta)}=1$

Solution:

Step 1: Write everything in terms of sin and cos on one side

$\dfrac{1}{\cos^2(\theta)}(1-\cos^2(\theta))=\dfrac{\sin^2(\theta)}{1-\sin^2(\theta)}$

Step 2: Reduce if possible.

$\dfrac{1}{\cos^2(\theta)}-\dfrac{\cos^2(\theta)}{\cos^2(\theta)}=\dfrac{\sin^2(\theta)}{1-\sin^2(\theta)}$

Step 3: Apply known identities $\sin^2(\theta)-1=\cos^2(\theta)$

$\sec^2(\theta)-1=\dfrac{\sin^2(\theta)}{1-\sin^2(\theta)}$

Step 4: If both sides of the equation match, it is an identity.

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