Basic Trigonometric Functions and Identities

Learning Outcomes

Convert angle measures between degrees and radians
Recognize the triangular and circular definitions of the basic trigonometric functions
Write the basic trigonometric identities

Radian Measure

To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, radians are a more natural measurement because they are related directly to the unit circle, a circle with radius 1. The radian measure of an angle is defined as follows. Given an angle $\theta$ , let $s$ be the length of the corresponding arc on the unit circle (Figure 1). We say the angle corresponding to the arc of length 1 has radian measure 1.

An image of a circle. At the exact center of the circle there is a point. From this point, there is one line segment that extends horizontally to the right a point on the edge of the circle and another line segment that extends diagonally upwards and to the right to another point on the edge of the circle. These line segments have a length of 1 unit. The curved segment on the edge of the circle that connects the two points at the end of the line segments is labeled “s”. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label “theta = s radians”.

Figure 1. The radian measure of an angle $\theta$ is the arc length $s$ of the associated arc on the unit circle.

Since an angle of 360° corresponds to the circumference of a circle, or an arc of length $2\pi$ , we conclude that an angle with a degree measure of 360° has a radian measure of $2\pi$ . Similarly, we see that 180° is equivalent to $\pi$ radians. The table below shows the relationship between common degree and radian values.

Table 1. Common Angles Expressed in Degrees and Radians
Degrees	Radians	Degrees	Radians
0	0	120	$2\pi/3$
30	$\pi/6$	135	$3\pi/4$
45	$\pi/4$	150	$5\pi/6$
60	$\pi/3$	180	$\pi$
90	$\pi/2$

Though the common radian and degree equivalents in the table above are worth memorizing, if you can’t remember them, don’t forget that we have a conversion formula!

Recall: radian and degree conversions

Converting between Radians and Degrees

Because degrees and radians both measure angles, we need to be able to convert between them. We can easily do so using a proportion.

\frac{θ}{180} = \frac{θ^{R}}{π}

$\frac{\theta }{180}=\frac{{\theta }^{R}}{\pi }$

This proportion shows that the measure of angle $\theta$ in degrees divided by 180 equals the measure of angle $\theta$ in radians divided by $\pi .$ Or, phrased another way, degrees is to 180 as radians is to $\pi$ .

\frac{Degrees}{180} = \frac{Radians}{π}

$\frac{\text{Degrees}}{180}=\frac{\text{Radians}}{\pi }$

Converting between Radians and Degrees

To convert between degrees and radians, use the proportion

\frac{θ}{180} = \frac{θ^{R}}{π}

$\frac{\theta }{180}=\frac{{\theta }^{R}}{\pi }$

Example: Converting between Radians and Degrees

Express 225° using radians.
Express $\dfrac{5\pi}{3}$ rad using degrees.

Show Solution

Use the fact that 180° is equivalent to $\pi$ radians as a conversion factor: $1=\dfrac{\pi \, \text{rad}}{180^{\circ}}=\dfrac{180^{\circ}}{\pi \, \text{rad}}$ .

$225^{\circ}=225^{\circ}·\dfrac{\pi }{180^{\circ}}=\dfrac{5\pi }{4}\text{ rad}$
$\dfrac{5\pi }{3}\text{ rad}$ = $\dfrac{5\pi }{3}·\dfrac{180^{\circ}}{\pi }=300^{\circ}$

Try It

Express 210° using radians. Express $\dfrac{11\pi}{6}$ rad using degrees.

Hint

$\pi$ radians is equal to $180^{\circ}$ .

Show Solution

$\dfrac{7\pi}{6}$ rad; 330°

Try It

The Six Basic Trigonometric Functions

Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.

To define the trigonometric functions, first consider the unit circle centered at the origin and a point $P=(x,y)$ on the unit circle. Let $\theta$ be an angle with an initial side that lies along the positive $x$ -axis and with a terminal side that is the line segment $OP$ . An angle in this position is said to be in standard position (Figure 2). We can then define the values of the six trigonometric functions for $\theta$ in terms of the coordinates $x$ and $y$ .

An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally upwards and to the right to another point on the edge of the circle. This point is labeled “P = (x, y)”. These line segments have a length of 1 unit. From the point “P”, there is a dotted vertical line that extends downwards until it hits the x axis and thus the horizontal line segment. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label “theta”.

Figure 2. The angle $\theta$ is in standard position. The values of the trigonometric functions for $\theta$ are defined in terms of the coordinates $x$ and $y$ .

Definition

Let $P=(x,y)$ be a point on the unit circle centered at the origin $O$ . Let $\theta$ be an angle with an initial side along the positive $x$ -axis and a terminal side given by the line segment $OP$ . The trigonometric functions are then defined as

\begin{array}{cccc} \sin θ = y & \csc θ = \frac{1}{y} \\ \cos θ = x & \sec θ = \frac{1}{x} \\ \tan θ = \frac{y}{x} & \cot θ = \frac{x}{y} \end{array}

$\begin{array}{cccc}\sin \theta =y & & & \csc \theta =\large{\frac{1}{y}} \normalsize \\ \cos \theta =x & & & \sec \theta =\large{\frac{1}{x}} \normalsize \\ \tan \theta =\large{\frac{y}{x}} \normalsize & & & \cot \theta =\large{\frac{x}{y}} \end{array}$

If $x=0$ , then $\sec \theta$ and $\tan \theta$ are undefined. If $y=0$ , then $\cot \theta$ and $\csc \theta$ are undefined.

We can see that for a point $P=(x,y)$ on a circle of radius $r$ with a corresponding angle $\theta$ , the coordinates $x$ and $y$ satisfy

\begin{matrix} \cos θ = \frac{x}{r} \\ x = r \cos θ \\ \sin θ = \frac{y}{r} \\ y = r \sin θ \end{matrix}

$\begin{array}{c} \cos \theta =\large{\frac{x}{r}} \\ x=r \cos \theta \\ \\ \sin \theta =\large{\frac{y}{r}} \\ y=r \sin \theta \end{array}$

The values of the other trigonometric functions can be expressed in terms of $x, \, y$ , and $r$ (Figure 3).

Figure 3. For a point $P=(x,y)$ on a circle of radius $r$ , the coordinates $x$ and $y$ satisfy $x=r \cos \theta$ and $y=r \sin \theta$ .

The table below shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of $\sin \theta$ and $\cos \theta$ .

Values of $\sin \theta$ and $\cos \theta$ at Major Angles $\theta$ in the First Quadrant
$\theta$	$\sin \theta$	$\cos \theta$
0	0	1
$\large{\frac{\pi}{6}}$	$\large{\frac{1}{2}}$	$\large{\frac{\sqrt{3}}{2}}$
$\large{\frac{\pi}{4}}$	$\large{\frac{\sqrt{2}}{2}}$	$\large{\frac{\sqrt{2}}{2}}$
$\large{\frac{\pi}{3}}$	$\large{\frac{\sqrt{3}}{2}}$	$\large{\frac{1}{2}}$
$\large{\frac{\pi}{2}}$	1	0

Example: Evaluating Trigonometric Functions

Evaluate each of the following expressions.

$\sin \Big(\large\frac{2\pi}{3}\Big)$
$\cos \Big(-\large\frac{5\pi}{6}\Big)$
$\tan \Big(\large\frac{15\pi}{4}\Big)$

Show Solution

On the unit circle, the angle $\theta =\large\frac{2\pi}{3}$ corresponds to the point $\Big(-\large\frac{1}{2}, \frac{\sqrt{3}}{2}\Big)$ . Therefore, $\sin \Big(\large\frac{2\pi}{3}\Big) \normalsize = y = \large\frac{\sqrt{3}}{2}$ .

Figure 4. Unit circle depiction of $\theta$
An angle $\theta =-\large\frac{5\pi}{6}$ corresponds to a revolution in the negative direction, as shown. Therefore, $\cos \Big(-\large\frac{5\pi}{6}\Big)$ $=x=-\large\frac{\sqrt{3}}{2}$ .

Figure 5. Unit circle depiction of $\theta$
An angle $\theta =\large\frac{15\pi}{4} \normalsize = 2\pi +\large\frac{7\pi}{4}$ . Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of $\large\frac{7\pi}{4}$ corresponds to the point $\Big(\large\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\Big)$ , we can conclude that $\tan \Big(\large\frac{15\pi}{4}\Big)\normalsize =\large\frac{y}{x}$ $=-1$ .

Figure 6. Unit circle depiction of $\theta$

Try It

Evaluate $\cos \left(\dfrac{3\pi}{4}\right)$ and $\sin \left(\dfrac{−\pi}{6}\right)$ .

Hint

Show Solution

$\cos \left(\dfrac{3\pi}{4}\right)=\dfrac{−\sqrt{2}}{2}; \, \sin\left(\dfrac{−\pi}{6}\right)=\dfrac{-1}{2}$

Try It

As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions evaluated at either of the acute angles of the triangle. Let $\theta$ be one of the acute angles. Let $A$ be the length of the adjacent leg, $O$ be the length of the opposite leg, and $H$ be the length of the hypotenuse. By inscribing the triangle into a circle of radius $H$ , as shown in Figure 7, we see that $A, \, H$ , and $O$ satisfy the following relationships with $\theta$ :

\begin{array}{cccc} \sin θ = \frac{O}{H} & \csc θ = \frac{H}{O} \\ \cos θ = \frac{A}{H} & \sec θ = \frac{H}{A} \\ \tan θ = \frac{O}{A} & \cot θ = \frac{A}{O} \end{array}

$\begin{array}{cccc}\sin \theta =\large \frac{O}{H} & & & \normalsize \csc \theta =\large \frac{H}{O} \\ \cos \theta =\large \frac{A}{H} & & & \sec \theta =\large \frac{H}{A} \\ \tan \theta =\large \frac{O}{A} & & & \cot \theta =\large \frac{A}{O} \end{array}$

Figure 7. By inscribing a right triangle in a circle, we can express the ratios of the side lengths in terms of the trigonometric functions evaluated at $\theta$ .

Example: Constructing a Wooden Ramp

A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is 4 ft from the ground and the angle between the ground and the ramp is to be $10^{\circ}$ , how long does the ramp need to be?

Show Solution

Let $x$ denote the length of the ramp. In the following image, we see that $x$ needs to satisfy the equation $\sin(10^{\circ})=\dfrac{4}{x}$ . Solving this equation for $x$ , we see that $x=\frac{4}{ \sin(10^{\circ})} \approx 23.035$ ft.

Figure 8. Sketch of the ramp and staircase.

Watch the following video to see the worked solution to Example: Constructing a Wooden Ramp

Closed Captioning and Transcript Information for Video

Try It

A house painter wants to lean a 20-ft ladder against a house. If the angle between the base of the ladder and the ground is to be $60^{\circ}$ , how far from the house should she place the base of the ladder?

Hint

Show Solution

Trigonometric Identities

A trigonometric identity is an equation involving trigonometric functions that is true for all angles $\theta$ for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next.

Trigonometric Identities

Reciprocal identities

\begin{array}{cccc} \tan θ = \frac{\sin θ}{\cos θ} & \cot θ = \frac{\cos θ}{\sin θ} \\ \csc θ = \frac{1}{\sin θ} & \sec θ = \frac{1}{\cos θ} \end{array}

$\begin{array}{cccc}\tan \theta =\large \frac{\sin \theta}{\cos \theta} & & & \cot \theta =\large \frac{\cos \theta}{\sin \theta} \\ \csc \theta =\large \frac{1}{\sin \theta} & & & \sec \theta =\large \frac{1}{\cos \theta} \end{array}$

Pythagorean identities

\sin^{2} θ + \cos^{2} θ = 1 1 + \tan^{2} θ = \sec^{2} θ 1 + \cot^{2} θ = \csc^{2} θ

$\sin^2 \theta +\cos^2 \theta =1\phantom{\rule{2em}{0ex}}1+\tan^2 \theta =\sec^2 \theta \phantom{\rule{2em}{0ex}}1+\cot^2 \theta =\csc^2 \theta$

Addition and subtraction formulas

\sin (α \pm β) = \sin α \cos β \pm \cos α \sin β

$\sin(\alpha \pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$

\cos (α \pm β) = \cos α \cos β \mp \sin α \sin β

$\cos(\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$

Double-angle formulas

\sin (2 θ) = 2 \sin θ \cos θ

$\sin(2\theta)=2\sin \theta \cos \theta$

\cos (2 θ) = 2 \cos^{2} θ - 1 = 1 - 2 \sin^{2} θ = \cos^{2} θ - \sin^{2} θ

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

Remember, solving a trigonometric equation is not very different from solving an algebraic equation.

Recall: Given a trigonometric equation, solve using algebra.

Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
Substitute the trigonometric expression with a single variable, such as $x$ or $u$ .
Solve the equation the same way an algebraic equation would be solved.
Substitute the trigonometric expression back in for the variable in the resulting expressions.
Solve for the angle.

Try It

Additionally, trigonometric equations will have an infinite number of solutions. If we need to find all possible solutions, then we must add $2\pi k$ , where $k$ is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is $2\pi :$

\sin θ = \sin (θ \pm 2 k π)

$\sin \theta =\sin \left(\theta \pm 2k\pi \right)$

Example: Solving Trigonometric Equations

For each of the following equations, use a trigonometric identity to find all solutions.

1. $1+\cos(2\theta)=\cos \theta$
2. $\sin(2\theta)=\tan \theta$

Show Solution

a. Using the double-angle formula for

\cos (2 θ)

$\cos(2\theta)$ , we see that

θ

$\theta$ is a solution of

$1+\cos(2\theta)=\cos \theta$

if and only if

$1+2\cos^2 \theta -1=\cos \theta$ ,

which is true if and only if

$2\cos^2 \theta -\cos \theta =0$ .

To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by $\cos \theta$ . The problem with dividing by $\cos \theta$ is that it is possible that $\cos \theta$ is zero. In fact, if we did divide both sides of the equation by $\cos \theta$ , we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that $\theta$ is a solution of this equation if and only if

$\cos \theta (2\cos \theta -1)=0$

Since $\cos \theta =0$ when

$\theta =\large \frac{\pi }{2}, \, \frac{\pi}{2} \normalsize \pm \pi, \, \large \frac{\pi}{2} \normalsize \pm 2\pi, \cdots$ ,

and $\cos \theta =1/2$ when

$\theta =\large \frac{\pi}{3}, \, \frac{\pi}{3} \normalsize \pm 2\pi, \cdots$ , or $\theta =-\large \frac{\pi}{3}, \, -\frac{\pi}{3} \normalsize \pm 2\pi, \cdots$ ,

we conclude that the set of solutions to this equation is

$\theta =\large \frac{\pi}{2} \normalsize +n\pi, \, \theta =\large \frac{\pi}{3} \normalsize +2n\pi$ , and $\theta =-\large \frac{\pi}{3}\normalsize +2n\pi, \, n=0, \pm 1, \pm 2,\cdots$

b. Using the double-angle formula for $\sin(2\theta)$ and the reciprocal identity for $\tan(\theta)$ , the equation can be written as

$2\sin \theta \cos \theta =\large \frac{\sin\theta}{\cos \theta}$

To solve this equation, we multiply both sides by $\cos \theta$ to eliminate the denominator, and say that if $\theta$ satisfies this equation, then $\theta$ satisfies the equation

$2\sin \theta \cos^2 \theta -\sin \theta =0$

However, we need to be a little careful here. Even if $\theta$ satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by $\cos \theta$ . However, if $\cos \theta =0$ , we cannot divide both sides of the equation by $\cos \theta$ . Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor $\sin \theta$ out of both terms on the left-hand side instead of dividing both sides of the equation by $\sin \theta$ . Factoring the left-hand side of the equation, we can rewrite this equation as

$\sin \theta (2\cos^2 \theta -1)=0$

Therefore, the solutions are given by the angles $\theta$ such that $\sin \theta =0$ or $\cos^2 \theta =1/2$ . The solutions of the first equation are $\theta =0, \pm \pi, \pm 2\pi, \cdots$ . The solutions of the second equation are $\theta =\pi /4, \, (\pi/4) \pm (\pi/2), \, (\pi/4) \pm \pi, \cdots$ . After checking for extraneous solutions, the set of solutions to the equation is

$\theta =n\pi$ and $\theta =\large \frac{\pi}{4}+\frac{n\pi}{2}, \, \normalsize n=0, \pm 1, \pm 2, \cdots$ .

Watch the following video to see the worked solution to Example: Solving Trigonometric Equations

Closed Captioning and Transcript Information for Video

Try It

Find all solutions to the equation $\cos(2\theta)=\sin \theta$ .

Hint

Show Solution

$\theta =\large \frac{3\pi}{2} \normalsize +2n\pi, \, \large \frac{\pi}{6} \normalsize +2n\pi, \, \large \frac{5\pi}{6} \normalsize +2n\pi$ for $n=0, \pm 1, \pm 2, \cdots$

Trigonometric identities are not only useful for solving trigonometric equations, they can also help us verify identities.

Recall: Given a trigonometric identity, verify that it is true.

Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.
Look for opportunities to factor expressions, square a binomial, or add fractions.
Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.
If these steps do not yield the desired result, try converting all terms to sines and cosines.

Example: Proving a Trigonometric Identity

Prove the trigonometric identity $1+\tan^2 \theta =\sec^2 \theta$ .

Show Solution

We start with the identity

\sin^{2} θ + \cos^{2} θ = 1

$\sin^2 \theta +\cos^2 \theta =1$

Dividing both sides of this equation by $\cos^2 \theta$ , we obtain

\frac{\sin^{2} θ}{\cos^{2} θ} + 1 = \frac{1}{\cos^{2} θ}

$\frac{\sin^2 \theta}{\cos^2 \theta}+1=\frac{1}{\cos^2 \theta}$

Since $\sin \theta / \cos \theta =\tan \theta$ and $1 / \cos \theta =\sec \theta$ , we conclude that

\tan^{2} θ + 1 = \sec^{2} θ

$\tan^2 \theta +1=\sec^2 \theta$ .

Try It

Prove the trigonometric identity $1+\cot^2 \theta =\csc^2 \theta$ .

Hint

Divide both sides of the identity $\sin^2 \theta + \cos^2 \theta =1$ by $\sin^2 \theta$ .

Module 1: Functions and Graphs