Learning Objectives

In this section you will:

7.3.1 – Find function values for sine and cosine for special angles.

7.3.1 – 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, and we found their sines and cosines using right triangles. We can also calculate sines and cosines of the special angles using the Pythagorean Identity.

Finding Sines and Cosines of [latex]\,45^{\circ}\,[/latex] Angles

First, we will look at angles of [latex]\,45^{\circ}\,[/latex] or [latex]\,\frac{\pi }{4},[/latex] as shown in (Figure). A [latex]\,45^{\circ}–45^{\circ}–90^{\circ}\,[/latex] 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 [latex]\,t=\frac{\pi }{4},[/latex] which is 45 degrees, the radius of the unit circle bisects the first quadrantal angle. This means the radius lies along the line [latex]\,y=x.\,[/latex] A unit circle has a radius equal to 1 so the right triangle formed below the line [latex]\,y=x\,[/latex] has sides [latex]\,x\,[/latex] and [latex]\,y\text{ }\left(y=x\right),[/latex] and radius = 1. See (Figure).

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

From the Pythagorean Theorem we get


We can then substitute [latex]\,y=x.[/latex]


Next we combine like terms.


And solving for [latex]\,x,[/latex] we get

[latex]$$\begin{array}{ccc}\hfill {x}^{2}& =& \frac{1}{2}\hfill \\ \hfill x& =& ±\frac{1}{\sqrt{2}}\hfill \end{array}$$[/latex]

In quadrant I, [latex]\,x=\frac{1}{\sqrt{2}}.[/latex]

At [latex]\,t=\frac{\pi }{4}\,[/latex] or 45 degrees,

[latex]$$\begin{array}{ccc}\hfill \left(x,y\right)& =& \left(x,x\right)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\hfill \\ \hfill x& =& \frac{1}{\sqrt{2}},y=\frac{1}{\sqrt{2}}\hfill \\ \hfill \text{cos }t& =& \frac{1}{\sqrt{2}},\text{sin }t=\frac{1}{\sqrt{2}}\hfill \end{array}$$[/latex]

If we then rationalize the denominators, we get

[latex]$$\begin{array}{ccc}\hfill \text{cos }t& =& \frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}\\ & =& \frac{\sqrt{2}}{2}\hfill \\ \text{sin }t& =& \frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}\hfill \\ & =& \frac{\sqrt{2}}{2}\hfill \end{array}$$[/latex]

Therefore, the [latex]\,\left(x,y\right)\,[/latex] coordinates of a point on a circle of radius [latex]\,1\,[/latex] at an angle of [latex]\,45^{\circ}\,[/latex] are [latex]\,\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right).[/latex]

Finding Sines and Cosines of [latex]\,30^{\circ}\,[/latex] and [latex]\,60^{\circ}\,[/latex] Angles

Next, we will find the cosine and sine at an angle of [latex]\,30^{\circ},[/latex] or [latex]\,\frac{\pi }{6}.\,[/latex] First, we will draw a triangle inside a circle with one side at an angle of [latex]\,30^{\circ},[/latex] and another at an angle of [latex]\,-30^{\circ},[/latex] as shown in (Figure). If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be [latex]\,60^{\circ},[/latex] as shown in (Figure).

Graph of a circle with 30-degree angle and negative 30-degree angle inscribed to form a triangle.
Figure 12.
Image of two 30/60/90 triangles back to back. Label for hypotenuse r and side y.
Figure 13.

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

[latex]\mathrm{sin}\left(\frac{\pi }{6}\right)=\frac{1}{2}r[/latex]

And since [latex]\,r=1\,[/latex] in our unit circle,

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

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

[latex]$$\begin{array}{cccc}\hfill {\mathrm{cos}}^{2}\left(\frac{\pi }{6}\right)+{\mathrm{sin}}^{2}\left(\frac{\pi }{6}\right)& =& 1\hfill & \\ \hfill {\mathrm{cos}}^{2}\left(\frac{\pi }{6}\right)+{\left(\frac{1}{2}\right)}^{2}& =& 1\hfill & \\ \hfill {\mathrm{cos}}^{2}\left(\frac{\pi }{6}\right)& =& \frac{3}{4}\hfill & \phantom{\rule{1em}{0ex}}\text{Use the square root property}.\hfill \\ \hfill \mathrm{cos}\left(\frac{\pi }{6}\right)& =& \frac{±\sqrt{3}}{±\sqrt{4}}=\frac{\sqrt{3}}{2}\hfill & \phantom{\rule{1em}{0ex}}\text{Since }y\text{ is positive, choose the positive root}.\hfill \end{array}$$[/latex]

The [latex]\,\left(x,y\right)\,[/latex] coordinates for the point on a circle of radius [latex]\,1\,[/latex] at an angle of [latex]\,30^{\circ}\,[/latex] are [latex]\,\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right).\,[/latex] At [latex]\,t=\frac{\pi }{3}\text{ (60°}\text{),}[/latex] the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, [latex]\,BAD,[/latex] as shown in (Figure). Angle [latex]\,A\,[/latex] has measure [latex]\,60°.\,[/latex] At point [latex]\,B,[/latex] we draw an angle [latex]\,ABC\,[/latex] with measure of [latex]\,60°.\,[/latex] We know the angles in a triangle sum to [latex]\,180°,[/latex] so the measure of angle [latex]\,C\,[/latex] is also [latex]\,60°.\,[/latex] Now we have an equilateral triangle. Because each side of the equilateral triangle [latex]\,ABC\,[/latex] 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 isosceles triangle inscribed that has been divided in half. The resulting triangle has a radius of 1 and a height of y. The two bases for the triangles each have a length of x.
Figure 13.

The measure of angle [latex]\,ABD\,[/latex] is 30°. Angle [latex]\,ABC\,[/latex] is double angle [latex]\,ABD,[/latex] so its measure is 60°. [latex]\,BD\,[/latex] is the perpendicular bisector of [latex]\,AC,[/latex] so it cuts [latex]\,AC\,[/latex] in half. This means that [latex]\,AD\,[/latex] is [latex]\,\frac{1}{2}\,[/latex] the radius, or [latex]\,\frac{1}{2}.\,[/latex] Notice that [latex]\,AD\,[/latex] is the x-coordinate of point [latex]\,B,[/latex] which is at the intersection of the 60° angle and the unit circle. This gives us a triangle [latex]\,BAD\,[/latex] with hypotenuse of 1 and side [latex]\,x\,[/latex] of length [latex]\,\frac{1}{2}.[/latex]

From the Pythagorean Theorem, we get


Substituting [latex]\,x=\frac{1}{2},[/latex] we get


Solving for [latex]\,y,[/latex] we get

[latex]$$\begin{array}{ccc}\hfill \frac{1}{4}+{y}^{2}& =& 1\hfill \\ \hfill {y}^{2}& =& 1-\frac{1}{4}\hfill \\ \hfill {y}^{2}& =& \frac{3}{4}\hfill \\ \hfill y& =& ±\frac{\sqrt{3}}{2}\hfill \end{array}$$[/latex]

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

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

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

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. (Figure) summarizes these values.

Angle [latex]0[/latex] [latex]\frac{\pi }{6},[/latex] or [latex]\,30^{\circ}[/latex] [latex]\frac{\pi }{4},[/latex] or [latex]\,45^{\circ}[/latex] [latex]\frac{\pi }{3},[/latex] or [latex]\,60^{\circ}[/latex] [latex]\frac{\pi }{2},[/latex] or [latex]\,90^{\circ}[/latex]
Cosine 1 [latex]\frac{\sqrt{3}}{2}[/latex] [latex]\frac{\sqrt{2}}{2}[/latex] [latex]\frac{1}{2}[/latex] 0
Sine 0 [latex]\frac{1}{2}[/latex] [latex]\frac{\sqrt{2}}{2}[/latex] [latex]\frac{\sqrt{3}}{2}[/latex] 1

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

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 [latex]\,\mathrm{cos}\left(30\right)\,[/latex] 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.

How To

Given an angle in radians, use a graphing calculator to find the cosine.

  1. If the calculator has degree mode and radian mode, set it to radian mode.
  2. Press the COS key.
  3. Enter the radian value of the angle and press the close-parentheses key “)”.
  4. Press ENTER.

Example 1 – Using a Graphing Calculator to Find Sine and Cosine

Evaluate [latex]\,\mathrm{cos}\left(\frac{5\pi }{3}\right)\,[/latex] using a graphing calculator or computer.

Enter the following keystrokes:

[latex]\text{COS}(\text{ }5\text{ }×\text{ }\pi \text{ }÷\text{ 3 ) ENTER}[/latex]

[latex]\mathrm{cos}\left(\frac{5\pi }{3}\right)=0.5[/latex]


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 sign of [latex]\,20°,[/latex] for example, by including the conversion factor to radians as part of the input:

[latex]\text{SIN}(\text{ 20 }×\text{ }\pi \text{ }÷\text{ 180 ) ENTER}[/latex]

Try It

Evaluate [latex]\,\mathrm{sin}\left(\frac{\pi }{3}\right).[/latex]

Show answer

approximately 0.866025403


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

  • 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.
  • Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known. See (Example 1).


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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