Unit Circle: Sine and Cosine Functions

Finding Function Values for the Sine and Cosine

To define our trigonometric functions, we begin by drawing a unit circle, a circle centered at the origin with radius 1, as shown in Figure 2. The angle (in radians) that $t$ intercepts forms an arc of length $s .$ Using the formula $s = r t,$ and knowing that $r = 1,$ we see that for a unit circle, $s = t .$

Recall that the x- and y-axes divide the coordinate plane into four quarters called quadrants. We label these quadrants to mimic the direction a positive angle would sweep. The four quadrants are labeled I, II, III, and IV.

For any angle $t,$ we can label the intersection of the terminal side and the unit circle as by its coordinates, $(x, y) .$ The coordinates $x$ and $y$ will be the outputs of the trigonometric functions $f (t) = \cos t$ and $f (t) = \sin t,$ respectively. This means $x = \cos t$ and $y = \sin t .$

Graph of a circle with angle t, radius of 1, and an arc created by the angle with length s. The terminal side of the angle intersects the circle at the point (x,y). — Figure 2 Unit circle where the central angle is $t$ radians

Defining Sine and Cosine Functions

Now that we have our unit circle labeled, we can learn how the $(x, y)$ coordinates relate to the arc length and angle. The sine function relates a real number $t$ to the y-coordinate of the point where the corresponding angle intercepts the unit circle. More precisely, the sine of an angle $t$ equals the y-value of the endpoint on the unit circle of an arc of length $t .$ In Figure 2, the sine is equal to $y .$ Like all functions, the sine function has an input and an output. Its input is the measure of the angle; its output is the y-coordinate of the corresponding point on the unit circle.

The cosine function of an angle $t$ equals the x-value of the endpoint on the unit circle of an arc of length $t .$ In Figure 3, the cosine is equal to $x .$

Illustration of an angle t, with terminal side length equal to 1, and an arc created by angle with length t. The terminal side of the angle intersects the circle at the point (x,y), which is equivalent to (cos t, sin t). — Figure 3

Because it is understood that sine and cosine are functions, we do not always need to write them with parentheses: $\sin t$ is the same as $\sin (t)$ and $\cos t$ is the same as $\cos (t) .$ Likewise, $\cos^{2} t$ is a commonly used shorthand notation for ${(\cos (t))}^{2} .$ Be aware that many calculators and computers do not recognize the shorthand notation. When in doubt, use the extra parentheses when entering calculations into a calculator or computer.

Example 1

Finding Function Values for Sine and Cosine

Point $P$ is a point on the unit circle corresponding to an angle of $t,$ as shown in Figure 4. Find $\cos (t)$ and $sin (t) .$

Graph of a circle with angle t, radius of 1, and a terminal side that intersects the circle at the point (1/2, square root of 3 over 2). — Figure 4

Solution

We know that $\cos t$ is the x-coordinate of the corresponding point on the unit circle and $\sin t$ is the y-coordinate of the corresponding point on the unit circle. So:

\begin{array}{l} \begin{array}{l} x = \cos t = \frac{1}{2} \end{array} \\ y = \sin t = \frac{\sqrt{3}}{2} \end{array}

Finding Sines and Cosines of Angles on an Axis

For quadrantral angles, the corresponding point on the unit circle falls on the x- or y-axis. In that case, we can easily calculate cosine and sine from the values of $x$ and $y .$

Example 2

Calculating Sines and Cosines along an Axis

Find $\cos (90°)$ and $sin (90°) .$

Solution

Moving $90°$ counterclockwise around the unit circle from the positive x-axis brings us to the top of the circle, where the $(x, y)$ coordinates are (0, 1), as shown in Figure 6.

Graph of a circle with angle t, radius of 1, and a terminal side that intersects the circle at the point (0,1). — Figure 6

Using our definitions of cosine and sine,

\begin{array}{l} x = \cos t = \cos (90°) = 0 \\ y = \sin t = \sin (90°) = 1 \end{array}

The cosine of 90° is 0; the sine of 90° is 1.

The Pythagorean Identity

Now that we can define sine and cosine, we will learn how they relate to each other and the unit circle. Recall that the equation for the unit circle is $x^{2} + y^{2} = 1.$ Because $x = \cos t$ and $y = \sin t,$ we can substitute for $x$ and $y$ to get $\cos^{2} t + \sin^{2} t = 1.$ This equation, $\cos^{2} t + \sin^{2} t = 1,$ is known as the Pythagorean Identity. See Figure 7.

Graph of an angle t, with a point (x,y) on the unit circle. And equation showing the equivalence of 1, x^2 + y^2, and cos^2 t + sin^2 t. — Figure 7

We can use the Pythagorean Identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.

Example 3

Finding a Cosine from a Sine or a Sine from a Cosine

If $\sin (t) = \frac{3}{7}$ and $t$ is in the second quadrant, find $\cos (t) .$

Solution

If we drop a vertical line from the point on the unit circle corresponding to $t,$ we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem. See Figure 8.

Graph of a unit circle with an angle that intersects the circle at a point with the y-coordinate equal to 3/7. — Figure 8

Substituting the known value for sine into the Pythagorean Identity,

\begin{array}{l} \cos^{2} (t) + \sin^{2} (t) = 1 \\ \cos^{2} (t) + \frac{9}{49} = 1 \\ \cos^{2} (t) = \frac{40}{49} \\ cos (t) = \pm \sqrt{\frac{40}{49}} = \pm \frac{\sqrt{40}}{7} = \pm \frac{2 \sqrt{10}}{7} \end{array}

Because the angle is in the second quadrant, we know the x-value is a negative real number, so the cosine is also negative. So $cos (t) = - \frac{2 \sqrt{10}}{7}$

Finding Sines and Cosines of Special Angles

We have already learned some properties of the special angles, such as the conversion from radians to degrees. We can also calculate sines and cosines of the special angles using the Pythagorean Identity and our knowledge of triangles.

Finding Sines and Cosines of 45° Angles

First, we will look at angles of $45°$ or $\frac{π}{4},$ as shown in Figure 9 . A $45° - 45° - 90°$ triangle is an isosceles triangle, so the x- and y-coordinates of the corresponding point on the circle are the same. Because the x- and y-values are the same, the sine and cosine values will also be equal.

Graph of 45 degree angle inscribed within a circle with radius of 1. Equivalence between point (x,y) and (x,x) shown. — Figure 9

At $t = \frac{π}{4}$ , which is 45 degrees, the radius of the unit circle bisects the first quadrantal angle. This means the radius lies along the line $y = x .$ A unit circle has a radius equal to 1. So, the right triangle formed below the line $y = x$ has sides $x$ and $y (y = x),$ and a radius = 1. See Figure 10.

Graph of circle with pi/4 angle inscribed and a radius of 1. — Figure 10

From the Pythagorean Theorem we get

x^{2} + y^{2} = 1

Substituting $y = x,$ we get

x^{2} + x^{2} = 1

Combining like terms we get

2 x^{2} = 1

And solving for $x,$ we get

\begin{matrix} x^{2} = \frac{1}{2} \\ x = \pm \frac{1}{\sqrt{2}} \end{matrix}

In quadrant I, $x = \frac{1}{\sqrt{2}} .$

At $t = \frac{π}{4}$ or 45 degrees,

\begin{array}{l} (x, y) = (x, x) = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \\ x = \frac{1}{\sqrt{2}}, y = \frac{1}{\sqrt{2}} \\ \cos t = \frac{1}{\sqrt{2}}, \sin t = \frac{1}{\sqrt{2}} \end{array}

If we then rationalize the denominators, we get

\begin{array}{l} \cos t = \frac{1}{\sqrt{2}} \frac{\sqrt{2}}{\sqrt{2}} \\ = \frac{\sqrt{2}}{2} \\ \sin t = \frac{1}{\sqrt{2}} \frac{\sqrt{2}}{\sqrt{2}} \\ = \frac{\sqrt{2}}{2} \end{array}

Therefore, the $(x, y)$ coordinates of a point on a circle of radius $1$ at an angle of $45°$ are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) .$

Finding Sines and Cosines of 30° and 60° Angles

Next, we will find the cosine and sine at an angle of $30°,$ or $\frac{π}{6}$ . First, we will draw a triangle inside a circle with one side at an angle of $30°,$ and another at an angle of $−30°,$ as shown in Figure 11. If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be $60°,$ as shown in Figure 12.

Graph of a circle with 30 degree angle and negative 30 degree angle inscribed to form a trangle. — Figure 11

Image of two 30/60/90 triangles back to back. Label for hypoteneuse r and side y. — Figure 12

Because all the angles are equal, the sides are also equal. The vertical line has length $2 y,$ and since the sides are all equal, we can also conclude that $r = 2 y$ or $y = \frac{1}{2} r .$ Since $\sin t = y$ ,

\sin (\frac{π}{6}) = \frac{1}{2} r

And since $r = 1$ in our unit circle,

\begin{array}{l} \sin (\frac{π}{6}) = \frac{1}{2} (1) \\ = \frac{1}{2} \end{array}

Using the Pythagorean Identity, we can find the cosine value.

\begin{array}{l} \cos^{2} \frac{π}{6} + \sin^{2} (\frac{π}{6}) = 1 \\ \cos^{2} (\frac{π}{6}) + {(\frac{1}{2})}^{2} = 1 \\ \cos^{2} (\frac{π}{6}) = \frac{3}{4} & Use the square root property . \\ \cos (\frac{π}{6}) = \frac{\pm \sqrt{3}}{\pm \sqrt{4}} = \frac{\sqrt{3}}{2} & Since y is positive, choose the positive root . \end{array}

The $(x, y)$ coordinates for the point on a circle of radius $1$ at an angle of $30°$ are $(\frac{\sqrt{3}}{2}, \frac{1}{2}) .$ At $t = \frac{π}{3}$ (60°), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, $B A D,$ as shown in Figure 13. Angle $A$ has measure $60° .$ At point $B,$ we draw an angle $A B C$ with measure of $60° .$ We know the angles in a triangle sum to $180°,$ so the measure of angle $C$ is also $60° .$ Now we have an equilateral triangle. Because each side of the equilateral triangle $A B C$ is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1.

Graph of circle with an isoceles triangle inscribed. — Figure 13

The measure of angle $A B D$ is 30°. So, if double, angle $A B C$ is 60°. $B D$ is the perpendicular bisector of $A C,$ so it cuts $A C$ in half. This means that $A D$ is $\frac{1}{2}$ the radius, or $\frac{1}{2} .$ Notice that $A D$ is the x-coordinate of point $B,$ which is at the intersection of the 60° angle and the unit circle. This gives us a triangle $B A D$ with hypotenuse of 1 and side $x$ of length $\frac{1}{2} .$

From the Pythagorean Theorem, we get

x^{2} + y^{2} = 1

Substituting $x = \frac{1}{2},$ we get

{(\frac{1}{2})}^{2} + y^{2} = 1

Solving for $y,$ we get

\begin{array}{l} \frac{1}{4} + y^{2} = 1 \\ y^{2} = 1 - \frac{1}{4} \\ y^{2} = \frac{3}{4} \\ y = \pm \frac{\sqrt{3}}{2} \end{array}

Since $t = \frac{π}{3}$ has the terminal side in quadrant I where the y-coordinate is positive, we choose $y = \frac{\sqrt{3}}{2},$ the positive value.

At $t = \frac{π}{3}$ (60°), the $(x, y)$ coordinates for the point on a circle of radius $1$ at an angle of $60°$ are $(\frac{1}{2}, \frac{\sqrt{3}}{2}),$ so we can find the sine and cosine.

\begin{array}{l} (x, y) = (\frac{1}{2}, \frac{\sqrt{3}}{2}) \\ x = \frac{1}{2}, y = \frac{\sqrt{3}}{2} \\ \cos t = \frac{1}{2}, \sin t = \frac{\sqrt{3}}{2} \end{array}

We have now found the cosine and sine values for all of the most commonly encountered angles in the first quadrant of the unit circle. Table 1 summarizes these values.

Table 1 ..
Angle	0	$\frac{π}{6},$ or 30	$\frac{π}{4},$ or 45°	$\frac{π}{3},$ or 60°	$\frac{π}{2},$ or 90°
Cosine	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0
Sine	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1

Figure 14 shows the common angles in the first quadrant of the unit circle.

Graph of a quarter circle with angles of 0, 30, 45, 60, and 90 degrees inscribed. Equivalence of angles in radians shown. Points along circle are marked. — Figure 14

Using a Calculator to Find Sine and Cosine

To find the cosine and sine of angles other than the special angles, we turn to a computer or calculator. Be aware: Most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When we evaluate $\cos (30)$ on our calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode.

Example 4

Using a Graphing Calculator to Find Sine and Cosine

Evaluate $\cos (\frac{5 π}{3})$ using a graphing calculator or computer.

Solution

Enter the following keystrokes:

COS $(5 \times π \div 3)$ ENTER

\cos (\frac{5 π}{3}) = 0.5

Analysis

We can find the cosine or sine of an angle in degrees directly on a calculator with degree mode. For calculators or software that use only radian mode, we can find the sine of $20°,$ for example, by including the conversion factor to radians as part of the input:

SIN $(20 \times π \div 180)$ ENTER

Identifying the Domain and Range of Sine and Cosine Functions

Now that we can find the sine and cosine of an angle, we need to discuss their domains and ranges. What are the domains of the sine and cosine functions? That is, what are the smallest and largest numbers that can be inputs of the functions? Because angles smaller than 0 and angles larger than $2 π$ can still be graphed on the unit circle and have real values of $x, y,$ and $r,$ there is no lower or upper limit to the angles that can be inputs to the sine and cosine functions. The input to the sine and cosine functions is the rotation from the positive x-axis, and that may be any real number.

What are the ranges of the sine and cosine functions? What are the least and greatest possible values for their output? We can see the answers by examining the unit circle, as shown in Figure 15. The bounds of the x-coordinate are $[−1, 1] .$ The bounds of the y-coordinate are also $[−1, 1] .$ Therefore, the range of both the sine and cosine functions is $[−1, 1] .$

Finding Reference Angles

We have discussed finding the sine and cosine for angles in the first quadrant, but what if our angle is in another quadrant? For any given angle in the first quadrant, there is an angle in the second quadrant with the same sine value. Because the sine value is the y-coordinate on the unit circle, the other angle with the same sine will share the same y-value, but have the opposite x-value. Therefore, its cosine value will be the opposite of the first angle’s cosine value.

Likewise, there will be an angle in the fourth quadrant with the same cosine as the original angle. The angle with the same cosine will share the same x-value but will have the opposite y-value. Therefore, its sine value will be the opposite of the original angle’s sine value.

As shown in Figure 16, angle $α$ has the same sine value as angle $t;$ the cosine values are opposites. Angle $β$ has the same cosine value as angle $t;$ the sine values are opposites.

\begin{array}{l} \sin (t) = \sin (α) & and & \cos (t) = - \cos (α) \\ \sin (t) = - \sin (β) & and & \cos (t) = \cos (β) \end{array}

Graph of two side by side circles. First graph has circle with angle t and angle alpha with radius r. Second graph has circle with angle t and angle beta inscribed with radius r. — Figure 16

Recall that an angle’s reference angle is the acute angle, $t,$ formed by the terminal side of the angle $t$ and the horizontal axis. A reference angle is always an angle between $0$ and $90°,$ or $0$ and $\frac{π}{2}$ radians. As we can see from Figure 17, for any angle in quadrants II, III, or IV, there is a reference angle in quadrant I.

Four side by side graphs. First graph shows an angle of t in quadrant 1 in it's normal position. Second graph shows an angle of t in quadrant 2 due to a rotation of pi minus t. Third graph shows an angle of t in quadrant 3 due to a rotation of t minus pi. Fourth graph shows an angle of t in quadrant 4 due to a rotation of two pi minus t. — Figure 17

Example 5

Finding a Reference Angle

Find the reference angle of $225°$ as shown in Figure 18.

Graph of circle with 225 degree angle inscribed. — Figure 18

Solution

Because $225°$ is in the third quadrant, the reference angle is

| (180° −225°) | = | - 45° | = 45°

Using Reference Angles

Now let’s take a moment to reconsider the Ferris wheel introduced at the beginning of this section. Suppose a rider snaps a photograph while stopped twenty feet above ground level. The rider then rotates three-quarters of the way around the circle. What is the rider’s new elevation? To answer questions such as this one, we need to evaluate the sine or cosine functions at angles that are greater than 90 degrees or at a negative angle. Reference angles make it possible to evaluate trigonometric functions for angles outside the first quadrant. They can also be used to find $(x, y)$ coordinates for those angles. We will use the reference angle of the angle of rotation combined with the quadrant in which the terminal side of the angle lies.

Using Reference Angles to Evaluate Trigonometric Functions

We can find the cosine and sine of any angle in any quadrant if we know the cosine or sine of its reference angle. The absolute values of the cosine and sine of an angle are the same as those of the reference angle. The sign depends on the quadrant of the original angle. The cosine will be positive or negative depending on the sign of the x-values in that quadrant. The sine will be positive or negative depending on the sign of the y-values in that quadrant.

Example 6

Using Reference Angles to Find Sine and Cosine

ⓐ Using a reference angle, find the exact value of $\cos (150°)$ and $sin (150°) .$
ⓑ Using the reference angle, find $\cos \frac{5 π}{4}$ and $\sin \frac{5 π}{4} .$

Solution

ⓐ 150° is located in the second quadrant. The angle it makes with the x-axis is 180° − 150° = 30°, so the reference angle is 30°.
This tells us that 150° has the same sine and cosine values as 30°, except for the sign. We know that

$\cos (30°) = \frac{\sqrt{3}}{2} and \sin (30°) = \frac{1}{2} .$

Since 150° is in the second quadrant, the x-coordinate of the point on the circle is negative, so the cosine value is negative. The y-coordinate is positive, so the sine value is positive.

$\cos (150°) = - \frac{\sqrt{3}}{2} and \sin (150°) = \frac{1}{2}$
ⓑ $\frac{5 π}{4}$ is in the third quadrant. Its reference angle is $\frac{5 π}{4} - π = \frac{π}{4} .$ The cosine and sine of $\frac{π}{4}$ are both $\frac{\sqrt{2}}{2} .$ In the third quadrant, both $x$ and $y$ are negative, so:
$\cos \frac{5 π}{4} = - \frac{\sqrt{2}}{2} and \sin \frac{5 π}{4} = - \frac{\sqrt{2}}{2}$

Using Reference Angles to Find Coordinates

Now that we have learned how to find the cosine and sine values for special angles in the first quadrant, we can use symmetry and reference angles to fill in cosine and sine values for the rest of the special angles on the unit circle. They are shown in Figure 19. Take time to learn the $(x, y)$ coordinates of all of the major angles in the first quadrant.

Graph of unit circle with angles in degrees, angles in radians, and points along the circle inscribed. — Figure 19 Special angles and coordinates of corresponding points on the unit circle

In addition to learning the values for special angles, we can use reference angles to find $(x, y)$ coordinates of any point on the unit circle, using what we know of reference angles along with the identities

\begin{matrix} x = \cos t & y = \sin t \end{matrix}

First we find the reference angle corresponding to the given angle. Then we take the sine and cosine values of the reference angle, and give them the signs corresponding to the y- and x-values of the quadrant.

Example 7

Using the Unit Circle to Find Coordinates

Find the coordinates of the point on the unit circle at an angle of $\frac{7 π}{6} .$

Solution

We know that the angle $\frac{7 π}{6}$ is in the third quadrant.

First, let’s find the reference angle by measuring the angle to the x-axis. To find the reference angle of an angle whose terminal side is in quadrant III, we find the difference of the angle and $π .$

\frac{7 π}{6} - π = \frac{π}{6}

Next, we will find the cosine and sine of the reference angle:

\cos (\frac{π}{6}) = \frac{\sqrt{3}}{2} \sin (\frac{π}{6}) = \frac{1}{2}

We must determine the appropriate signs for x and y in the given quadrant. Because our original angle is in the third quadrant, where both $x$ and $y$ are negative, both cosine and sine are negative.

\begin{array}{l} \cos (\frac{7 π}{6}) = - \frac{\sqrt{3}}{2} \\ \sin (\frac{7 π}{6}) = - \frac{1}{2} \end{array}

Now we can calculate the $(x, y)$ coordinates using the identities $x = \cos θ$ and $y = \sin θ .$

The coordinates of the point are $(- \frac{\sqrt{3}}{2}, - \frac{1}{2})$ on the unit circle.

Key Equations

..
Cosine	$\cos t = x$
Sine	$\sin t = y$
Pythagorean Identity	$\cos^{2} t + \sin^{2} t = 1$

Key Concepts

Finding the function values for the sine and cosine begins with drawing a unit circle, which is centered at the origin and has a radius of 1 unit.
Using the unit circle, the sine of an angle $t$ equals the y-value of the endpoint on the unit circle of an arc of length $t$ whereas the cosine of an angle $t$ equals the x-value of the endpoint. See Example 1.
The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. See Example 2.
When the sine or cosine is known, we can use the Pythagorean Identity to find the other. The Pythagorean Identity is also useful for determining the sines and cosines of special angles. See Example 3.
Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known. See Example 4.
The domain of the sine and cosine functions is all real numbers.
The range of both the sine and cosine functions is $[- 1, 1] .$
The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle.
The signs of the sine and cosine are determined from the x- and y-values in the quadrant of the original angle.
An angle’s reference angle is the size angle, $t,$ formed by the terminal side of the angle $t$ and the horizontal axis. See Example 5.
Reference angles can be used to find the sine and cosine of the original angle. See Example 6.
Reference angles can also be used to find the coordinates of a point on a circle. See Example 7.

Section Exercises

Verbal

Exercise 1

Describe the unit circle.

Solution

The unit circle is a circle of radius 1 centered at the origin.

Exercise 2

What do the x- and y-coordinates of the points on the unit circle represent?

Exercise 3

Discuss the difference between a coterminal angle and a reference angle.

Solution

Coterminal angles are angles that share the same terminal side. A reference angle is the size of the smallest acute angle, $t,$ formed by the terminal side of the angle $t$ and the horizontal axis.

Exercise 4

Explain how the cosine of an angle in the second quadrant differs from the cosine of its reference angle in the unit circle.

Exercise 5

Explain how the sine of an angle in the second quadrant differs from the sine of its reference angle in the unit circle.

Solution

The sine values are equal.

Algebraic

For the following exercises, use the given sign of the sine and cosine functions to find the quadrant in which the terminal point determined by $t$ lies.

Exercise 6

$sin (t) < 0$ and $cos (t) < 0$

Exercise 7

$sin (t) > 0$ and $\cos (t) > 0$

Solution

I

Exercise 8

$\sin (t) > 0$ and $\cos (t) < 0$

Exercise 9

$\sin (t) < 0$ and $\cos (t) > 0$

Solution

IV

For the following exercises, find the exact value of each trigonometric function.

Exercise 10

$\sin \frac{π}{2}$

Exercise 11

$\sin \frac{π}{3}$

Solution

$\frac{\sqrt{3}}{2}$

Exercise 12

$\cos \frac{π}{2}$

Exercise 13

$\cos \frac{π}{3}$

Solution

$\frac{1}{2}$

Exercise 14

$\sin \frac{π}{4}$

Exercise 15

$\cos \frac{π}{4}$

Solution

$\frac{\sqrt{2}}{2}$

Exercise 16

$\sin \frac{π}{6}$

Exercise 17

$\sin π$

Solution

0

Exercise 18

$\sin \frac{3 π}{2}$

Exercise 19

$\cos π$

Solution

−1

Exercise 20

$\cos 0$

Exercise 21

$\cos \frac{π}{6}$

Solution

$\frac{\sqrt{3}}{2}$

Exercise 22

$\sin 0$

Numeric

For the following exercises, state the reference angle for the given angle.

Exercise 23

$240°$

Solution

$60°$

Exercise 24

$- 170°$

Exercise 25

$100°$

Solution

$80°$

Exercise 26

$- 315°$

Exercise 27

$135°$

Solution

$45°$

Exercise 28

$\frac{5 π}{4}$

Exercise 29

$\frac{2 π}{3}$

Solution

$\frac{π}{3}$

Exercise 30

$\frac{5 π}{6}$

Exercise 31

$\frac{- 11 π}{3}$

Solution

$\frac{π}{3}$

Exercise 32

$\frac{- 7 π}{4}$

Exercise 33

$\frac{- π}{8}$

Solution

$\frac{π}{8}$

For the following exercises, find the reference angle, the quadrant of the terminal side, and the sine and cosine of each angle. If the angle is not one of the special angles on the unit circle, use a calculator and round to three decimal places.

Exercise 34

$225°$

Exercise 35

$300°$

Solution

$60°,$ Quadrant IV, $sin (300°) = - \frac{\sqrt{3}}{2}, \cos (300°) = \frac{1}{2}$

Exercise 36

$320°$

Exercise 37

$135°$

Solution

$45°,$ Quadrant II, $sin (135°) = \frac{\sqrt{2}}{2},$ $\cos (135°) = - \frac{\sqrt{2}}{2}$

Exercise 38

$210°$

Exercise 39

$120°$

Solution

$60°,$ Quadrant II, $sin (120°) = \frac{\sqrt{3}}{2},$ $\cos (120°) = - \frac{1}{2}$

Exercise 40

$250°$

Exercise 41

$150°$

Solution

$30°,$ Quadrant II, $sin (150°) = \frac{1}{2},$ $\cos (150°) = - \frac{\sqrt{3}}{2}$

Exercise 42

$\frac{5 π}{4}$

Exercise 43

$\frac{7 π}{6}$

Solution

$\frac{π}{6},$ Quadrant III, $sin (\frac{7 π}{6}) = - \frac{1}{2},$ $cos (\frac{7 π}{6}) = - \frac{\sqrt{3}}{2}$

Exercise 44

$\frac{5 π}{3}$

Exercise 45

$\frac{3 π}{4}$

Solution

$\frac{π}{4},$ Quadrant II, $sin (\frac{3 π}{4}) = \frac{\sqrt{2}}{2},$ $\cos (\frac{3 π}{4}) = - \frac{\sqrt{2}}{2}$

Exercise 46

$\frac{4 π}{3}$

Exercise 47

$\frac{2 π}{3}$

Solution

$\frac{π}{3},$ Quadrant II, $sin (\frac{2 π}{3}) = \frac{\sqrt{3}}{2},$ $\cos (\frac{2 π}{3}) = - \frac{1}{2}$

Exercise 48

$\frac{5 π}{6}$

Exercise 49

$\frac{7 π}{4}$

Solution

$\frac{π}{4},$ Quadrant IV, $sin (\frac{7 π}{4}) = - \frac{\sqrt{2}}{2},$ $cos (\frac{7 π}{4}) = \frac{\sqrt{2}}{2}$

For the following exercises, find the requested value.

Exercise 50

If $cos (t) = \frac{1}{7}$ and $t$ is in the 4^th quadrant, find $sin (t) .$

Exercise 51

If $cos (t) = \frac{2}{9}$ and $t$ is in the 1^st quadrant, find $sin (t) .$

Solution

$\frac{\sqrt{77}}{9}$

Exercise 52

If $sin (t) = \frac{3}{8}$ and $t$ is in the 2^nd quadrant, find $cos (t) .$

Exercise 53

If $sin (t) = - \frac{1}{4}$ and $t$ is in the 3^rd quadrant, find $cos (t) .$

Solution

$- \frac{\sqrt{15}}{4}$

Exercise 54

Find the coordinates of the point on a circle with radius 15 corresponding to an angle of $220° .$

Exercise 55

Find the coordinates of the point on a circle with radius 20 corresponding to an angle of $120° .$

Solution

$(- 10, 10 \sqrt{3})$

Exercise 56

Find the coordinates of the point on a circle with radius 8 corresponding to an angle of $\frac{7 π}{4} .$

Exercise 57

Find the coordinates of the point on a circle with radius 16 corresponding to an angle of $\frac{5 π}{9} .$

Solution

$(- 2.778, 15.757)$

Exercise 58

State the domain of the sine and cosine functions.

Exercise 59

State the range of the sine and cosine functions.

Solution

$[- 1, 1]$

Graphical

For the following exercises, use the given point on the unit circle to find the value of the sine and cosine of $t .$

Exercise 60

Exercise 61

Graph of circle with angle of t inscribed. Point of (negative square root of 3 over 2, 1/2) is at intersection of terminal side of angle and edge of circle.

Solution

$\sin t = \frac{1}{2}, \cos t = - \frac{\sqrt{3}}{2}$

Exercise 62

Graph of circle with angle of t inscribed. Point of (1/2, negative square root of 3 over 2) is at intersection of terminal side of angle and edge of circle.

Exercise 63

Graph of circle with angle of t inscribed. Point of (negative square root of 2 over 2, negative square root of 2 over 2) is at intersection of terminal side of angle and edge of circle.

Solution

$\sin t = - \frac{\sqrt{2}}{2}, \cos t = - \frac{\sqrt{2}}{2}$

Exercise 64

Graph of circle with angle of t inscribed. Point of (1/2, square root of 3 over 2) is at intersection of terminal side of angle and edge of circle.

Exercise 65

Solution

$\sin t = \frac{\sqrt{3}}{2}, \cos t = - \frac{1}{2}$

Exercise 66

Exercise 67

Graph of circle with angle of t inscribed. Point of (square root of 2 over 2, negative square root of 2 over 2) is at intersection of terminal side of angle and edge of circle.

Solution

$\sin t = - \frac{\sqrt{2}}{2}, \cos t = \frac{\sqrt{2}}{2}$

Exercise 68

Graph of circle with angle of t inscribed. Point of (1,0) is at intersection of terminal side of angle and edge of circle.

Exercise 69

Solution

$\sin t = 0, \cos t = - 1$

Exercise 70

Graph of circle with angle of t inscribed. Point of (0.111,0.994) is at intersection of terminal side of angle and edge of circle.

Exercise 71

Graph of circle with angle of t inscribed. Point of (0.803,-0.596 is at intersection of terminal side of angle and edge of circle.

Solution

$\sin t = - 0.596, \cos t = 0.803$

Exercise 72

Graph of circle with angle of t inscribed. Point of (negative square root of 2 over 2, square root of 2 over 2) is at intersection of terminal side of angle and edge of circle.

Exercise 73

Graph of circle with angle of t inscribed. Point of (square root of 3 over 2, 1/2) is at intersection of terminal side of angle and edge of circle.

Solution

$\sin t = \frac{1}{2}, \cos t = \frac{\sqrt{3}}{2}$

Exercise 74

Exercise 75

Solution

$\sin t = - \frac{1}{2}, \cos t = \frac{\sqrt{3}}{2}$

Exercise 76

Graph of circle with angle of t inscribed. Point of (0, -1) is at intersection of terminal side of angle and edge of circle.

Exercise 77

Graph of circle with angle of t inscribed. Point of (-0.649, 0.761) is at intersection of terminal side of angle and edge of circle.

Solution

$\sin t = 0.761, \cos t = - 0.649$

Exercise 78

Graph of circle with angle of t inscribed. Point of (-0.948, -0.317) is at intersection of terminal side of angle and edge of circle.

Exercise 79

Solution

$\sin t = 1, \cos t = 0$

Technology

For the following exercises, use a graphing calculator to evaluate.

Exercise 80

$\sin \frac{5 π}{9}$

Exercise 81

$\cos \frac{5 π}{9}$

Solution

−0.1736

Exercise 82

$\sin \frac{π}{10}$

Exercise 83

$\cos \frac{π}{10}$

Solution

0.9511

Exercise 84

$\sin \frac{3 π}{4}$

Exercise 85

$\cos \frac{3 π}{4}$

Solution

−0.7071

Exercise 86

$\sin 98°$

Exercise 87

$\cos 98°$

Solution

−0.1392

Exercise 88

$\cos 310°$

Exercise 89

$\sin 310°$

Solution

−0.7660

Extensions

For the following exercises, evaluate.

Exercise 90

$\sin (\frac{11 π}{3}) \cos (\frac{- 5 π}{6})$

Exercise 91

$\sin (\frac{3 π}{4}) \cos (\frac{5 π}{3})$

Solution

$\frac{\sqrt{2}}{4}$

Exercise 92

$\sin (- \frac{4 π}{3}) \cos (\frac{π}{2})$

Exercise 93

$\sin (\frac{- 9 π}{4}) \cos (\frac{- π}{6})$

Solution

$- \frac{\sqrt{6}}{4}$

Exercise 94

$\sin (\frac{π}{6}) \cos (\frac{- π}{3})$

Exercise 95

$\sin (\frac{7 π}{4}) \cos (\frac{- 2 π}{3})$

Solution

$\frac{\sqrt{2}}{4}$

Exercise 96

$\cos (\frac{5 π}{6}) \cos (\frac{2 π}{3})$

Exercise 97

$\cos (\frac{- π}{3}) \cos (\frac{π}{4})$

Solution

$\frac{\sqrt{2}}{4}$

Exercise 98

$\sin (\frac{- 5 π}{4}) \sin (\frac{11 π}{6})$

Exercise 99

$\sin (π) \sin (\frac{π}{6})$

Solution

0

Real-World Applications

For the following exercises, use this scenario: A child enters a carousel that takes one minute to revolve once around. The child enters at the point $(0, 1),$ that is, on the due north position. Assume the carousel revolves counter clockwise.

Exercise 100

What are the coordinates of the child after 45 seconds?

Exercise 101

What are the coordinates of the child after 90 seconds?

Solution

$(0, - 1)$

Exercise 102

What is the coordinates of the child after 125 seconds?

Exercise 103

When will the child have coordinates $(0.707, –0.707)$ if the ride lasts 6 minutes? (There are multiple answers.)

Solution

37.5 seconds, 97.5 seconds, 157.5 seconds, 217.5 seconds, 277.5 seconds, 337.5 seconds

Exercise 104

When will the child have coordinates $(−0.866, −0.5)$ if the ride last 6 minutes?