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Unit Circle: Sine and Cosine Functions

Find function values for the sine and cosine of

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 . The angle (in radians) that \(t\) intercepts forms an arc of length \(s.\) Using the formula \(s=rt,\) 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 \(y=\sin \ t.\)

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

Try it.

Find \(\cos (90^{\circ})\) and \(\text{sin}(90^{\circ}).\)

Solution

Moving \(90^{\circ}\) 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 .

Using our definitions of cosine and sine,

\[\begin{array}{l}x=\cos \ t=\cos (90^{\circ})=0 \\ y=\sin \ t=\sin (90^{\circ})=1\end{array}\]

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

Condensed — the full section is in OpenStax Precalculus 2e.

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.

First, we will look at angles of \(45^{\circ}\) or \(\frac{\pi }{4},\) as shown in . A \(45^{\circ}-45^{\circ}-90^{\circ}\) 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.

At \(t=\frac{\pi }{4}\), 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 .

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

\[\begin{array}{l}{x}^{2}=\frac{1}{2} \\ x=\pm \frac{1}{\sqrt{2}}\end{array}\]

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

At \(t=\frac{\pi }{4}\) 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 \(45^{\circ}\) are \((\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}).\)

Condensed — the full section is in OpenStax Precalculus 2e.

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 \(2\pi\) 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 . 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 , angle \(\alpha \\) has the same sine value as angle \(\ t;\) the cosine values are opposites. Angle \(\beta \\) has the same cosine value as angle \(t;\) the sine values are opposites.

\[\begin{array}{lll}\sin (t)=\ \sin (\alpha ) & \text{and} & \cos (t)=-\cos (\alpha ) \\ \sin (t)=-\sin (\beta ) & \text{and} & \cos (t)=\ \cos (\beta )\end{array}\]

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 \(90^{\circ},\) or \(0\) and \(\frac{\pi }{2}\) radians. As we can see from , for any angle in quadrants II, III, or IV, there is a reference angle in quadrant I.

Example

Try it.

Find the reference angle of \(225^{\circ}\) as shown in .

Solution

Because \(225^{\circ}\) is in the third quadrant, the reference angle is

\[|(180^{\circ}-225^{\circ})|=|-45^{\circ}|=45^{\circ}\]

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.

Condensed — the full section is in OpenStax Precalculus 2e.

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 .
  • The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. See .
  • 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 .
  • Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known. See .
  • 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 .
  • Reference angles can be used to find the sine and cosine of the original angle. See .
  • Reference angles can also be used to find the coordinates of a point on a circle. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Point \(\ P\\) is a point on the unit circle corresponding to an angle of \(\ t,\) as shown in . Find \(\ \cos (t)\) and \(\text{sin}(t).\)

    Разкрийте отговора

    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:

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

  2. A certain angle \(t\) corresponds to a point on the unit circle at \((-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\) as shown in . Find \(\cos \ t\) and \(\sin \ t.\)

    Разкрийте отговора

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

  3. Find \(\cos (90^{\circ})\) and \(\text{sin}(90^{\circ}).\)

    Разкрийте отговора

    Moving \(90^{\circ}\) 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 .

    Using our definitions of cosine and sine,

    \[\begin{array}{l}x=\cos \ t=\cos (90^{\circ})=0 \\ y=\sin \ t=\sin (90^{\circ})=1\end{array}\]

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

  4. Find cosine and sine of the angle \(\pi .\)

    Разкрийте отговора

    \(\cos (\pi )=-1,\) \(\sin (\pi )=0\)

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

    Разкрийте отговора

    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 .

    Substituting the known value for sine into the Pythagorean Identity,

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

    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 \[\text{cos}(t)=-\frac{2\sqrt{10}}{7}\]

  6. If \(\cos (t)=\frac{24}{25}\) and \(t\) is in the fourth quadrant, find \(\text{sin}(t).\)

    Разкрийте отговора

    \(\sin (t)=-\frac{7}{25}\)

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

    Разкрийте отговора

    Enter the following keystrokes:

    COS \((5\times \pi \div 3)\) ENTER

    \[\cos (\frac{5\pi }{3})=0.5\]

  8. Evaluate \(\sin (\frac{\pi }{3}).\)

    Разкрийте отговора

    approximately 0.866025403

  9. Find the reference angle of \(225^{\circ}\) as shown in .

    Разкрийте отговора

    Because \(225^{\circ}\) is in the third quadrant, the reference angle is

    \[|(180^{\circ}-225^{\circ})|=|-45^{\circ}|=45^{\circ}\]

  10. Find the reference angle of \(\frac{5\pi }{3}.\)

    Разкрийте отговора

    \(\frac{\pi }{3}\)

    1. ⓐ Using a reference angle, find the exact value of \(\cos (150^{\circ})\) and \(\text{sin}(150^{\circ}).\)
    2. ⓑ Using the reference angle, find \(\cos \ \frac{5\pi }{4}\) and \(\sin \ \frac{5\pi }{4}.\)
    Разкрийте отговора
    1. ⓐ 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

      \[\cos (30^{\circ})=\frac{\sqrt{3}}{2}\ \text{and}\ \sin (30^{\circ})=\frac{1}{2}.\]

      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.

      \[\cos (150^{\circ})=-\frac{\sqrt{3}}{2}\ \text{and}\ \sin (150^{\circ})=\frac{1}{2}\]
    2. ⓑ \(\frac{5\pi }{4}\) is in the third quadrant. Its reference angle is \(\frac{5\pi }{4}-\pi =\frac{\pi }{4}.\) The cosine and sine of \(\frac{\pi }{4}\) are both \(\frac{\sqrt{2}}{2}.\) In the third quadrant, both \(x\) and \(y\) are negative, so: \[\cos \ \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}\ \text{and}\ \sin \ \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}\]
    1. ⓐ Use the reference angle of \(315^{\circ}\) to find \(\cos (315^{\circ})\) and \(\sin (315^{\circ}).\)
    2. ⓑ Use the reference angle of \(-\frac{\pi }{6}\) to find \(\cos (-\frac{\pi }{6})\) and \(\sin (-\frac{\pi }{6}).\)
    Разкрийте отговора

    1. ⓐ \(\text{cos}(315^{\circ})=\frac{\sqrt{2}}{2},\text{ sin}(315^{\circ})=\frac{-\sqrt{2}}{2}\)
    2. ⓑ \(\cos (-\frac{\pi }{6})=\frac{\sqrt{3}}{2},\sin (-\frac{\pi }{6})=-\frac{1}{2}\)

  13. Find the coordinates of the point on the unit circle at an angle of \(\frac{7\pi }{6}.\)

    Разкрийте отговора

    We know that the angle \(\frac{7\pi }{6}\) 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 \(\pi .\)

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

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

    \[\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}\ \sin (\frac{\pi }{6})=\frac{1}{2}\]

    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.

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

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

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

  14. Find the coordinates of the point on the unit circle at an angle of \(\frac{5\pi }{3}.\)

    Разкрийте отговора

    \((\frac{1}{2},-\frac{\sqrt{3}}{2})\)

  15. Describe the unit circle.

    Разкрийте отговора

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

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

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

    Разкрийте отговора

    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.

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

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

    Разкрийте отговора

    The sine values are equal.

  20. \(\text{sin}(t)<0\) and \(\text{cos}(t)<0\)

  21. \(\text{sin}(t)>0\) and \(\cos (t)>0\)

    Разкрийте отговора

    I

  22. \(\sin (t)>0\) and \(\cos (t)<0\)

  23. \(\sin (t)<0\) and \(\cos (t)>0\)

    Разкрийте отговора

    IV

  24. \(\sin \ \frac{\pi }{2}\)

  25. \(\sin \ \frac{\pi }{3}\)

    Разкрийте отговора

    \(\frac{\sqrt{3}}{2}\)

  26. \(\cos \ \frac{\pi }{2}\)

  27. \(\cos \ \frac{\pi }{3}\)

    Разкрийте отговора

    \(\frac{1}{2}\)

  28. \(\sin \ \frac{\pi }{4}\)

  29. \(\cos \ \frac{\pi }{4}\)

    Разкрийте отговора

    \(\frac{\sqrt{2}}{2}\)

  30. \(\sin \ \frac{\pi }{6}\)

  31. \(\sin \ \pi\)

    Разкрийте отговора

    0

  32. \(\sin \ \frac{3\pi }{2}\)

  33. \(\cos \ \pi\)

    Разкрийте отговора

    −1

  34. \(\cos \ 0\)

  35. \(\cos \ \frac{\pi }{6}\)

    Разкрийте отговора

    \(\frac{\sqrt{3}}{2}\)

  36. \(\sin \ 0\)

  37. \(240^{\circ}\)

    Разкрийте отговора

    \(60^{\circ}\)

  38. \(-170^{\circ}\)

  39. \(100^{\circ}\)

    Разкрийте отговора

    \(80^{\circ}\)

  40. \(-315^{\circ}\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\pi
pi
Ratio of a circle's circumference to its diameter, 3.14159…
\sin,\ \cos,\ \tan
sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.
^\circ
degrees
1/360 of a full turn. 180° = π radians.
\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\theta
theta
The usual name for an angle.
\arcsin,\ \sin^{-1}
inverse sine
The angle whose sine is the given value (and likewise arccos, arctan).

How to: Unit Circle: Sine and Cosine Functions

  1. Find function values for the sine and cosine of
  2. Identify the domain and range of sine and cosine functions.
  3. Use reference angles to evaluate trigonometric functions.
  4. The sine of
  5. The cosine of
  6. Substitute the known value of
  7. Solve for
  8. Choose the solution with the appropriate sign for the

Questions people ask

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

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Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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