Trigonometry

With our basic coordinate geometry, we are now able to locate objects in space. As we start discussing how those objects are moving through space we are going to need a description of direction which will require angles. To that end, let us look at some basic concepts from trigonometry that we will find useful.

Angles

In this section, we will look at the two most common measures of angle. We will look at degrees because that is what we are most used to dealing with in our everyday lives. We will look at radians because those are the units used in C++.
An angle is a measure of rotation. This is usually represented graphically as the amount of rotation on a ray from some initial position to some terminal position.

Degrees

One degree is defined as 1/360th of a revolution. In other words, a complete revolution will be 360 degrees.

Radians

One radian is defined as the angle that would sweep out an arc length of 1 on the unit circle. This connection between the angle and the arc length is the main reason we use it exclusively in Calculus.

Converting Between Degrees and Radians

We know that one full rotation contains $360^{\circ}$. We also know that a full rotation would sweep out the entire circumference of the unit circle. Since the circumference of the unit circle is $C=2\pi$, we then know that one complete rotation is $2\pi$ radians. Now we simply set up the proportion,

(1)

\begin{align} \frac{\mbox{Degrees}}{\mbox{Radians}}=\frac{360}{2\pi} \end{align}

Solving this proportion for either degrees or radians gives us the following two conversion formulas

(2)

\begin{eqnarray} \mbox{Degrees}=\mbox{Radians} \times \frac{180}{\pi} \\ \mbox{Radians}=\mbox{Degrees} \times \frac{\pi}{180} \end{eqnarray}

Example
Convert $\frac{\pi}{36}$ to degrees

Solution

$\left ( \frac{\pi}{36} \right ) \left ( \frac{ 180^{\circ}}{\pi} \right )= 5^{\circ}$

Example
Convert $315^{\circ}$ to radians

Solution

$\left ( \frac{315^{\circ}}{1} \right ) \left ( \frac{\pi}{ 180^{\circ}}\right ) = \frac{7\pi}{4}$

The Trigonometric Functions

Trigonometric functions are a way in which we relate angles to either the xy-coordinate system or to the sides of a right triangle. The trigonometric functions are sine, cosine, tangent, secant, cosecant, and cotangent.

Trigonometry on the Right Triangle

A right triangle is a triangle where one of the angles is a right angle(i.e. $90^{\circ} \mbox{ or } \frac{\pi}{2}$). We refer to the two sides that make up the right angle as the legs and the side opposite the right angle the hypotenuse.
We can define the values of the trigonometric function for an angle $0^\circ<\theta<90^\circ$ of a right triangle as the ratio of particular sides on that triangle as follows.

(3)

\begin{align} \sin \theta = \frac{\mbox{Opposite}}{\mbox{Hypotenuse}} & \cos \theta =\frac{\mbox{Adjacent}}{\mbox{Hypotenuse}} & \tan \theta =\frac{\mbox{Opposite}}{\mbox{Adjacent}}\\ & & \\ \sec \theta = \frac{\mbox{Hypotenuse}}{\mbox{Adjacent}} & \csc \theta = \frac{\mbox{Hypotenuse}}{\mbox{Opposite}} & \cot \theta = \frac{\mbox{Adjacent}}{\mbox{Opposite}}\\ \end{align}

(Insert Exercise with a triangle here)
In general

Trigonometry on a Circle

The one limitation to defining the values of the trigonometric functions on the right triangle is the fact that $0^\circ<\theta<90^\circ$. Now we will connect the values of trigonometric functions to the points on a circle in the xy-plane and therefore allow any angle of $\theta$. The image below shows a circle of radius r centered at the origin. The right triangle has been superimposed in order to show how the trigonometric values from the right triangle relate to the points on the circle.

From this image we can see that the values of the trigonometric functions are as follows

$\sin \theta = \frac{y}{r}$	$\cos \theta =\frac{x}{r}$	$\tan \theta =\frac{y}{x}$
$\sec \theta = \frac{r}{x}$	$\csc \theta = \frac{r}{y}$	$\cot \theta =\frac{x}{y}$

The most common circle to use is the unit circle. The unit circle is a circle of radius 1. In that case the previous formulae simplify to

$\sin \theta = y$	$\cos \theta =x$	$\tan \theta =\frac{y}{x}$
$\sec \theta = \frac{1}{x}$	$\csc \theta = \frac{1}{y}$	$\cot \theta =\frac{x}{y}$

Trigonometric functions are different than algebraic functions in that there is no finite formula that can tell you how to compute their value for any arbitrary angle. Functions with this property are called transcendental functions. There are several methods for approximating these functions that can be implemented algorithmically, such as CORDIC, Taylor Polynomials, Chebyshev Polynomials, and other so-called minimax approximations. There are some common angles for which the values of the trigonometric functions can be determined exactly via geometric derivation. A compact list of these common angles in the first quadrant is given below.

$\theta$	$0^{\circ}$	$30^{\circ}$	$45^{\circ}$	$60^{\circ}$	$90^{\circ}$
$\sin (\theta)$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$
$\cos (\theta)$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$

Reference Angles

Due to the symmetry of the unit circle, it is possible to relate the values of the trigonometric functions for angles in quadrants II, III and IV to the values of the trigonometric functions for angles in quadrant I. This is done through what is called the reference angle. In the image below, you can see that there are four angles that make an angle of $\theta$ with the x-axis. Since the four intersection points on the unit circle, have various axis symmetries, the coordinates are the same in absolute value.

So, for example, the angles $150^{\circ} ,210^{\circ},$ and $330^{\circ}$ all have $30^{\circ}$ as their reference angle. So, knowing that $\sin \left ( 30^{\circ} \right ) = \frac{1}{2}$ and $\sin \left ( 30^{\circ} \right ) = \frac{sqrt{3}}{2}$ leads to the following,

$\theta$	$150^{\circ}$	$210^{\circ}$	$330^{\circ}$
$\sin (\theta)$	$\frac{1}{2}$	$-\frac{1}{2}$	$-\frac{1}{2}$
$\cos (\theta)$	$-\frac{\sqrt{3}}{2}$	$-\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{2}$

Example
Give the exact values of $\sin \left ( 300^{\circ} \right )$, $\cos \left ( 300^{\circ} \right )$, and $\tan \left ( 300^{\circ} \right )$,

Solution

$300^{\circ}$ has $60^{\circ}$ as its reference angle
We know that $\sin \left ( 60^{\circ} \right )=\frac{\sqrt{3}}{2}$ and $\cos \left ( 60^{\circ} \right )=\frac{1}{2}$.
We also know that $300^{\circ}$ is in the fouth quadrant, so on the unit circle $x>0$ and $y<0$, so
$\sin \left ( 300^{\circ} \right )=-\frac{\sqrt{3}}{2}$ and $\cos \left ( 300^{\circ} \right )=\frac{1}{2}$
Then by definition $\tan \left ( 300^{\circ} \right )= \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}$

The Inverse Trigonometric Functions

We have seen that trigonometric functions take an angle, in radians or degrees, as their input and output a dimensionless value that can be related to a right triangle or the unit circle. In many instances, we have that dimensionless value and are interested in the corresponding angle. This requires us to construct an inverse trigonometric function for each trigonometric function. There is only one catch. From algebra, we know that a function is invertible if and only if every element of the range corresponds to exactly one element in the domain. In other words, it must pass the horizontal line test. Due to their periodic nature, none of the trigonometric functions pass the horizontal line test and are therefore not invertible. All is not lost. If we restrict the domain of the trigonometric function, we can obtain a piece that is invertible.

ArcSine

If we restrict the function $y=\sin t$ to $-\frac{\pi}{2}\leq t \leq \frac{\pi}{2}$, it will pass the horizontal line test and therefore be invertible. The following graphs illustrate each.

The only problem with restricting the domain on $y=\sin t$ is that it restricts us to angles in the 1st and 4th quadrants. What about the angles in the 2nd and 3rd quadrants?

ArcCosine

For the same reason as above, we restrict $y=\cos t$ to $0 \leq t \leq \pi$ in order to obtain the function $y=\cos^{-1}t$

ArcTangent

Likewise we restrice $y=\tan t$ to $-\frac{\pi}{2} < t < \frac{\pi}{2}$ to obtain $y=\tan ^{-1}t$

In all three cases the restriction of the domain has the unintended effect of limiting our solutions when trying to use the inverse trigonometric functions to solve trigonometric equations. In these cases we need to use reference angles. More specifically, in order to obtain the other solution in the range $0 \leq \theta \leq 360^{\circ}$
you will need to do the following,

1. $180^{\circ}-\sin^{-1}(t)$ or $\pi-\sin^{-1}(t)$
2. $360^{\circ}-\cos^{-1}(t)$ or $2\pi-\cos^{-1}(t)$
3. $180^{\circ}+\tan^{-1}(t)$ or $\pi+\tan^{-1}(t)$

Example
Find all solutions for $\sin(\theta)= \frac{\sqrt{3}}{2}$ with $0 \leq \theta \leq 360^{\circ}$

Solution

We need to solve the equation $\sin(\theta)= \frac{\sqrt{3}}{2}$ by applying the inverse sine function to both sides
$\sin^{-1}(\sin(\theta))= \sin^{-1}(\frac{\sqrt{3}}{2})$
$\theta= \sin^{-1}(\frac{\sqrt{3}}{2})$ when restricted to $-90^{\circ} \leq \theta \leq 90^{\circ}$
$\theta=60^{\circ}$
And using reference angles $\theta=180^{\circ}-60^{\circ}=120^{\circ}$
So $\theta=60^{\circ},120^{\circ}$ for $0 \leq \theta \leq 360^{\circ}$

Example
Find all solutions for $\tan(\theta)= 1$ with $0 \leq \theta \leq 2\pi$

Solution

We need to solve the equation $\tan(\theta)= 1$ by applying the inverse tangent function to both sides
$\tan^{-1}(\tan(\theta))= \tan^{-1}(1)$
$\theta= \tan^{-1}(1)$ when restricted to $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$
$\theta=\frac{\pi}{4}$
And the solution in Quadrant III , $\theta=\frac{\pi}{4}+\pi=\frac{5\pi}{4}$
So $\theta=\frac{\pi}{4},\frac{5\pi}{4}$ for $0 \leq \theta \leq 2\pi$

Exercises

1. Convert $40^{\circ}$ to radians
2. Convert $305^{\circ}$ to radians
3. Convert $\frac{\pi}{20}$ to degrees
4. Convert $\frac{\pi}{100}$ to degrees
5. Approximate $\sin (22^{\circ})$ using a calculator
6. Approximate $\cos (147^{\circ})$ using a calculator
7. Find $\tan (90^{\circ})$
8. Approximate $\sin (2.4)$ using a calculator
9. Approximate $\cos(\frac{\pi}{8})$ using a calculator
10. Find $\tan(\pi)$
11. Find all solutions for $\sin(\theta)= \frac{1}{2}$ with $0 \leq \theta \leq 360^{\circ}$
12. Find all solutions for $\sin(\theta)= -\frac{\sqrt{3}}{2}$ with $0 \leq \theta \leq 360^{\circ}$
13. Find all solutions for $\cos(\theta)= -\frac{\sqrt{2}}{2}$ with $0 \leq \theta \leq 360^{\circ}$
14. Find all solutions for $\cos(\theta)= \frac{\sqrt{1}}{2}$ with $0 \leq \theta \leq 360^{\circ}$
15. Find all solutions for $\tan(\theta)= -\sqrt{3}$ with $0 \leq \theta \leq 360^{\circ}$
16. Find all solutions for $\tan(\theta)= 1$ with $0 \leq \theta \leq 360^{\circ}$
17. What are the smallest and largest values that $\sin \theta$ and $\cos \theta$ can attain? Explain why using the unit circle.
18. How does $\sin \left ( \frac{\pi}{4} \right )$ relate to $\sin \left ( \frac{3\pi}{4} \right )$ and why?
19. What is the domain of $f(x)=\sin (x)$ and $g(x)=\cos (x)$ assuming $x$ is measured in radians?
20. Using a calculator, approximate all solutions for $\sin(\theta)= \frac{1}{3}$ with $0 \leq \theta \leq 360^{\circ}$
21. Using a calculator, approximate all solutions for $\sin(\theta)= - 0.25$ with $0 \leq \theta \leq 360^{\circ}$
22. Using a calculator, approximate all solutions for $\cos(\theta)= -\frac{2}{3}$ with $0 \leq \theta \leq 360^{\circ}$
23. Using a calculator, approximate all solutions for $\cos(\theta)= 0.85$ with $0 \leq \theta \leq 360^{\circ}$
24. Using a calculator, approximate all solutions for $\tan(\theta)= - 3$ with $0 \leq \theta \leq 360^{\circ}$
25. Using a calculator, approximate all solutions for $\tan(\theta)= 2$ with $0 \leq \theta \leq 360^{\circ}$