Photo of a ferris wheel.
Figure 1. The Singapore Flyer is the world’s tallest Ferris wheel. (credit: ʺVibin JKʺ/Flickr)

Looking for a thrill? Then consider a ride on the Singapore Flyer, the world’s tallest Ferris wheel. Located in Singapore, the Ferris wheel soars to a height of 541 feet—a little more than a tenth of a mile! Described as an observation wheel, riders enjoy spectacular views as they travel from the ground to the peak and down again in a repeating pattern. In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.

7.4.1 – Finding Trigonometric Functions Using the Unit Circle

We have already defined the trigonometric functions in terms of right triangles. In this section, we will redefine them in terms of the unit circle. Recall that a unit circle is a circle centered at the origin with radius 1, as shown in (Figure). The angle (in radians) that [latex]\,t\,[/latex] intercepts forms an arc of length [latex]\,s.\,[/latex] Using the formula [latex]\,s=rt,[/latex] and knowing that [latex]\,r=1,[/latex] we see that for a unit circle, [latex]\,s=t.[/latex]

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 [latex]\,t,[/latex] we can label the intersection of the terminal side and the unit circle as by its coordinates, [latex]\,\left(x,y\right).\,[/latex] The coordinates [latex]\,x\,[/latex] and [latex]\,y\,[/latex] will be the outputs of the trigonometric functions [latex]\,f\left(t\right)=\mathrm{cos}\,t\,[/latex] and [latex]\,f\left(t\right)=\mathrm{sin}\,t,[/latex] respectively. This means [latex]\phantom{\rule{0.3em}{0ex}}x=\text{cos }t\phantom{\rule{0.3em}{0ex}}[/latex] and [latex]\phantom{\rule{0.3em}{0ex}}y=\text{sin }t.[/latex]

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

A unit circle has a center at [latex]\,\left(0,0\right)\,[/latex] and radius [latex]\,1.\,[/latex] In a unit circle, the length of the intercepted arc is equal to the radian measure of the central angle [latex]\,t.[/latex]

Let [latex]\,\left(x,y\right)\,[/latex] be the endpoint on the unit circle of an arc of arc length [latex]\,s.\,[/latex] The [latex]\,\left(x,y\right)\,[/latex] coordinates of this point can be described as functions of the angle.

Defining Sine and Cosine Functions from the Unit Circle

The sine function relates a real number [latex]\,t\,[/latex] to the y-coordinate of the point where the corresponding angle intercepts the unit circle. More precisely, the sine of an angle [latex]\,t\,[/latex] equals the y-value of the endpoint on the unit circle of an arc of length [latex]\,t.\,[/latex] In (Figure), the sine is equal to [latex]\,y.\,[/latex] 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 [latex]\,t\,[/latex] equals the x-value of the endpoint on the unit circle of an arc of length [latex]\,t.\,[/latex] In (Figure), the cosine is equal to [latex]\,x.[/latex]

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: [latex]\,\mathrm{sin}\,t\,[/latex] is the same as [latex]\,\mathrm{sin}\left(t\right)\,[/latex] and [latex]\,\mathrm{cos}t\,[/latex] is the same as [latex]\,\mathrm{cos}\left(t\right).\,[/latex] Likewise, [latex]\,{\mathrm{cos}}^{2}t\,[/latex] is a commonly used shorthand notation for [latex]\,{\left(\mathrm{cos}\left(t\right)\right)}^{2}.\,[/latex] 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.

Sine and Cosine Functions

If [latex]\,t\,[/latex] is a real number and a point [latex]\,\left(x,y\right)\,[/latex] on the unit circle corresponds to a central angle [latex]\,t,[/latex] then

[latex]\mathrm{cos}\,t=x[/latex]
[latex]\mathrm{sin}\,t=y[/latex]

How To

Given a point P [latex]\,\left(x,y\right)\,[/latex] on the unit circle corresponding to an angle of [latex]\,t,[/latex] find the sine and cosine.

  1. The sine of [latex]\,t\,[/latex] is equal to the y-coordinate of point [latex]\,P:\text{sin }t\text{ = }y.[/latex]
  2. The cosine of [latex]\,t\,[/latex] is equal to the x-coordinate of point [latex]\,P:\text{cos} t=x.[/latex]

Example 1 – Finding Function Values for Sine and Cosine

Point [latex]\,P\,[/latex] is a point on the unit circle corresponding to an angle of [latex]\,t,[/latex] as shown in (Figure). Find [latex]\,\mathrm{cos}\left(t\right)\,[/latex] and [latex]\,\text{sin}\left(t\right).[/latex]

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.

We know that [latex]\,\mathrm{cos}\,t\,[/latex] is the x-coordinate of the corresponding point on the unit circle and [latex]\,\mathrm{sin}\,t\,[/latex] is the y-coordinate of the corresponding point on the unit circle. So:

[latex]$$\begin{array}{ccc}\hfill x& =\mathrm{cos}\,t\hfill & =\frac{1}{2}\hfill \\ y& =\mathrm{sin}\,t\hfill & =\frac{\sqrt{3}}{2}\hfill \end{array}$$[/latex]

Try It

A certain angle [latex]\,t\,[/latex] corresponds to a point on the unit circle at [latex]\,\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\,[/latex] as shown in (Figure). Find [latex]\,\mathrm{cos}\,t\,[/latex] and [latex]\,\mathrm{sin}\,t.[/latex]

Graph of a circle with angle t, radius of 1, and a terminal side that intersects the circle at the point (negative square root of 2 over 2, square root of 2 over 2).
Figure 5.
Show answer

[latex]\mathrm{cos}\left(t\right)=-\frac{\sqrt{2}}{2},\mathrm{sin}\left(t\right)=\frac{\sqrt{2}}{2}[/latex]

 

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 [latex]\,x\,[/latex] and [latex]\,y.[/latex]

Example 2 – Calculating Sines and Cosines along an Axis

Find [latex]\,\text{cos}\left(90°\right)\,[/latex] and [latex]\,\text{sin}\left(90°\right).[/latex]

Moving [latex]\,90°\,[/latex] counterclockwise around the unit circle from the positive x-axis brings us to the top of the circle, where the [latex]\,\left(x,y\right)\,[/latex] coordinates are [latex]\,\left(0,1\right),[/latex] as shown in (Figure).

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.

We can then use our definitions of cosine and sine.

[latex]$$\begin{array}{cccc}\hfill x& =\text{cos }t\hfill & =\mathrm{cos}\left(90°\right)\hfill & =0\hfill \\ \hfill y& =\text{sin }t\hfill & =\mathrm{sin}\left(90°\right)\hfill & =1\hfill \end{array}$$[/latex]

The cosine of [latex]\,90°\,[/latex] is 0; the sine of [latex]\,90°\,[/latex] is 1.

Try It

Find cosine and sine of the angle [latex]\,\pi .[/latex]

Show answer

[latex]\mathrm{cos}\left(\pi \right)=-1,\mathrm{sin}\left(\pi \right)=0[/latex]

 

7.4.2 – 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), angle [latex]\,\alpha \,[/latex] has the same sine value as angle [latex]\,t;[/latex] the cosine values are opposites. Angle [latex]\,\beta \,[/latex] has the same cosine value as angle [latex]\,t;[/latex] the sine values are opposites.

[latex]$$\begin{array}{ccc}\mathrm{sin}\left(t\right)=\mathrm{sin}\left(\alpha \right)\hfill & \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}& \mathrm{cos}\left(t\right)=-\mathrm{cos}\left(\alpha \right)\hfill \\ \mathrm{sin}\left(t\right)=-\mathrm{sin}\left(\beta \right)\hfill & \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}& \mathrm{cos}\left(t\right)=\mathrm{cos}\left(\beta \right)\hfill \end{array}$$[/latex]
Graph of two side by side circles. First graph has circle with angle t and angle alpha with radius r. Angle t has its terminal side in Quadrant I whereas angle alpha has its terminal side in Quadrant II. Second graph has circle with angle t and angle beta inscribed with radius r. Angle t has its terminal side in Quadrant I whereas angle beta has its terminal side in Quadrant IV.
Figure 16.

Recall that an angle’s reference angle is the acute angle, [latex]\,t,[/latex] formed by the terminal side of the angle [latex]\,t\,[/latex] and the horizontal axis. A reference angle is always an angle between [latex]\,0\,[/latex] and [latex]\,90°,[/latex] or [latex]\,0\,[/latex] and [latex]\,\frac{\pi }{2}\,[/latex] radians. As we can see from (Figure), 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 its 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.

How To

Given an angle between [latex]\,0\,[/latex] and [latex]\,2\pi ,[/latex] find its reference angle.

  1. An angle in the first quadrant is its own reference angle.
  2. For an angle in the second or third quadrant, the reference angle is [latex]\,|\pi -t|\,[/latex] or [latex]\,|180°-t|.[/latex]
  3. For an angle in the fourth quadrant, the reference angle is [latex]\,2\pi -t\,[/latex] or [latex]\,360°-t.[/latex]
  4. If an angle is less than [latex]\,0\,[/latex] or greater than [latex]\,2\pi ,[/latex] add or subtract [latex]\,2\pi \,[/latex] as many times as needed to find an equivalent angle between [latex]\,0\,[/latex] and [latex]\,2\pi .[/latex]

Example 3 – Finding a Reference Angle

Find the reference angle of [latex]\,225°\,[/latex] as shown in (Figure).

Graph of circle with 225-degree angle inscribed.
Figure 17.

Because [latex]\,225°\,[/latex] is in the third quadrant, the reference angle is

[latex]|\left(180°-225°\right)|=|-45°|=45°[/latex]

Try It

Find the reference angle of [latex]\,\frac{5\pi }{3}.[/latex]

Show answer

[latex]\frac{\pi }{3}[/latex]

 

7.4.3 – 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 [latex]\,\left(x,y\right)\,[/latex] 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.

Using Reference Angles to Find Cosine and Sine

Angles have cosines and sines with the same absolute value as their reference angles. The sign (positive or negative) can be determined from the quadrant of the angle.

How To

Given an angle in standard position, find the reference angle, and the cosine and sine of the original angle.

  1. Measure the angle between the terminal side of the given angle and the horizontal axis. That is the reference angle.
  2. Determine the values of the cosine and sine of the reference angle.
  3. Give the cosine the same sign as the x-values in the quadrant of the original angle.
  4. Give the sine the same sign as the y-values in the quadrant of the original angle.

Example 4 – Using Reference Angles to Find Sine and Cosine

  1. Using a reference angle, find the exact value of [latex]\,\mathrm{cos}\left(150°\right)\,[/latex] and [latex]\,\text{sin}\left(150°\right).[/latex]
  2. Using the reference angle, find [latex]\,\mathrm{cos}\,\frac{5\pi }{4}\,[/latex] and [latex]\,\mathrm{sin}\,\frac{5\pi }{4}.[/latex]
  1. [latex]150°\,[/latex] is located in the second quadrant. The angle it makes with the x-axis is [latex]\,180°-150°=30°,[/latex] so the reference angle is [latex]\,30°.[/latex]

    This tells us that [latex]\,150°\,[/latex] has the same sine and cosine values as [latex]\,30°,[/latex] except for the sign.

    [latex]$$\begin{array}{ccc}\mathrm{cos}\left(30°\right)=\frac{\sqrt{3}}{2}& \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}& \mathrm{sin}\left(30°\right)=\frac{1}{2}\end{array}$$[/latex]

    Since [latex]\,150°\,[/latex] 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.

    [latex]$$\begin{array}{ccc}\mathrm{cos}\left(150°\right)=-\frac{\sqrt{3}}{2}& \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}& \mathrm{sin}\left(150°\right)=\frac{1}{2}\end{array}$$[/latex]
  2. [latex]\frac{5\pi }{4}\,[/latex] is in the third quadrant. Its reference angle is [latex]\,\frac{5\pi }{4}-\pi =\frac{\pi }{4}.\,[/latex] The cosine and sine of [latex]\,\frac{\pi }{4}\,[/latex] are both [latex]\,\frac{\sqrt{2}}{2}.\,[/latex] In the third quadrant, both [latex]\,x\,[/latex] and [latex]\,y\,[/latex] are negative, so:
    [latex]$$\begin{array}{ccc}\mathrm{cos}\phantom{\rule{0.03em}{0ex}}\frac{5\pi }{4}=-\frac{\sqrt{2}}{2}& \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}& \mathrm{sin}\phantom{\rule{0.03em}{0ex}}\frac{5\pi }{4}=-\frac{\sqrt{2}}{2}\end{array}$$[/latex]

Try It

  1. Use the reference angle of [latex]\,315°\,[/latex] to find [latex]\,\mathrm{cos}\left(315°\right)\,[/latex] and [latex]\,\mathrm{sin}\left(315°\right).[/latex]
  2. Use the reference angle of [latex]\,-\frac{\pi }{6}\,[/latex] to find [latex]\,\mathrm{cos}\left(-\frac{\pi }{6}\right)\,[/latex] and [latex]\,\mathrm{sin}\left(-\frac{\pi }{6}\right).[/latex]
Show answer
  1. [latex]\text{cos}\left(315°\right)=\frac{\sqrt{2}}{2}, \text{sin}\left(315°\right)=\frac{–\sqrt{2}}{2}[/latex]
  2. [latex]\text{cos}\left(-\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2},\mathrm{sin}\left(-\frac{\pi }{6}\right)=-\frac{1}{2}[/latex]

 

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). Take time to learn the [latex]\,\left(x,y\right)\,[/latex] 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 [latex]\,\left(x,y\right)\,[/latex] coordinates of any point on the unit circle, using what we know of reference angles along with the identities

[latex]$$\begin{array}{c}x=\text{cos }t\hfill \\ y=\text{sin }t\hfill \end{array}$$[/latex]

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.

How To

Given the angle of a point on a circle and the radius of the circle, find the [latex]\,\left(x,y\right)\,[/latex] coordinates of the point.

  1. Find the reference angle by measuring the smallest angle to the x-axis.
  2. Find the cosine and sine of the reference angle.
  3. Determine the appropriate signs for [latex]\,x\,[/latex] and [latex]\,y\,[/latex] in the given quadrant.

Example 5 – Using the Unit Circle to Find Coordinates

Find the coordinates of the point on the unit circle at an angle of [latex]\,\frac{7\pi }{6}.[/latex]

We know that the angle [latex]\,\frac{7\pi }{6}\,[/latex] 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 [latex]\,\pi .[/latex]

[latex]\frac{7\pi }{6}-\pi =\frac{\pi }{6}[/latex]

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

[latex]$$\begin{array}{cc}\mathrm{cos}\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\hfill & \phantom{\rule{1em}{0ex}}\mathrm{sin}\left(\frac{\pi }{6}\right)=\frac{1}{2}\hfill \end{array}$$[/latex]

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 [latex]\,x\,[/latex] and [latex]\,y\,[/latex] are negative, both cosine and sine are negative.

[latex]$$\begin{array}{ccc}\hfill \mathrm{cos}\left(\frac{7\pi }{6}\right)& =& -\frac{\sqrt{3}}{2}\hfill \\ \hfill \mathrm{sin}\left(\frac{7\pi }{6}\right)& =& -\frac{1}{2}\hfill \end{array}$$[/latex]

Now we can calculate the [latex]\,\left(x,y\right)\,[/latex] coordinates using the identities [latex]\,x=\mathrm{cos}\,\theta \,[/latex] and [latex]\,y=\mathrm{sin}\,\theta .[/latex]

The coordinates of the point are [latex]\,\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\,[/latex] on the unit circle.

Try It

Find the coordinates of the point on the unit circle at an angle of [latex]\,\frac{5\pi }{3}.[/latex]

Show answer

[latex]\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)[/latex]

 

Key Equations

Cosine [latex]\mathrm{cos}\,t=x[/latex]
Sine [latex]\mathrm{sin}\,t=y[/latex]
Pythagorean Identity [latex]{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1[/latex]

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 [latex]\,t\,[/latex] equals the y-value of the endpoint on the unit circle of an arc of length [latex]\,t\,[/latex] whereas the cosine of an angle [latex]\,t\,[/latex] 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).
  • 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, [latex]\,t,[/latex] formed by the terminal side of the angle [latex]\,t\,[/latex] and the horizontal axis. See (Example 3).
  • Reference angles can be used to find the sine and cosine of the original angle. See (Example 4).
  • Reference angles can also be used to find the coordinates of a point on a circle. See (Example 5).

Glossary

cosine function
the x-value of the point on a unit circle corresponding to a given angle
Pythagorean Identity
a corollary of the Pythagorean Theorem stating that the square of the cosine of a given angle plus the square of the sine of that angle equals 1
sine function
the y-value of the point on a unit circle corresponding to a given angle

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