49 Unit Circle

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]

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.

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]

[reveal-answer q=”fs-id1592628″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1592628″]

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][/hidden-answer]

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]

[reveal-answer q=”fs-id1614220″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1614220″]

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

[/hidden-answer]

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]

Calculating Sines and Cosines along an Axis

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

[reveal-answer q=”fs-id2204890″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id2204890″]

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).

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.[/hidden-answer]

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[latex]\,{x}^{2}+{y}^{2}=1.\,[/latex]Because[latex]\,x=\mathrm{cos}\,t\,[/latex]and[latex]\,y=\mathrm{sin}\,t,[/latex]we can substitute for[latex]\,x\,[/latex]and[latex]\,y\,[/latex]to get[latex]\,{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1.\,[/latex]This equation,[latex]\,{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1,[/latex]is known as the Pythagorean Identity. See (Figure).

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.

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

If[latex]\,\mathrm{sin}\left(t\right)=\frac{3}{7}\,[/latex]and[latex]\,t\,[/latex]is in the second quadrant, find[latex]\,\mathrm{cos}\left(t\right).[/latex]

[reveal-answer q=”fs-id1708122″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1708122″]

If we drop a vertical line from the point on the unit circle corresponding to[latex]\,t,[/latex]we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem. See (Figure).

Substituting the known value for sine into the Pythagorean Identity,

[latex]\begin{array}{ccc}\hfill {\text{cos}}^{2}\left(t\right)+{\mathrm{sin}}^{2}\left(t\right)& =& 1\hfill \\ \hfill {\text{cos}}^{2}\left(t\right)+\frac{9}{49}& =& 1\hfill \\ \hfill {\text{cos}}^{2}\left(t\right)& =& \frac{40}{49}\hfill \\ \hfill \text{cos}\left(t\right)& =& ±\sqrt{\frac{40}{49}}=±\frac{\sqrt{40}}{7}=±\frac{2\sqrt{10}}{7}\hfill \end{array}[/latex]

Because the angle is in the second quadrant, we know the x-value is a negative real number, so the cosine is also negative.

[latex]\text{cos}\left(t\right)=-\frac{2\sqrt{10}}{7}[/latex][/hidden-answer]

Finding Sines and Cosines of[latex]\,45°\,[/latex]Angles

First, we will look at angles of[latex]\,45°\,[/latex]or[latex]\,\frac{\pi }{4},[/latex]as shown in (Figure). A[latex]\,45°–45°–90°\,[/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.

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).

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

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

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

If we then rationalize the denominators, we get

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°\,[/latex]are[latex]\,\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right).[/latex]

Finding Sines and Cosines of[latex]\,30°\,[/latex]and[latex]\,60°\,[/latex]Angles

Next, we will find the cosine and sine at an angle of[latex]\,30°,[/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°,[/latex]and another at an angle of[latex]\,-30°,[/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°,[/latex]as shown in (Figure).

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]

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

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

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°\,[/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.

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

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.

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.

(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.

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 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.

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.

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]

[reveal-answer q=”fs-id2094916″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id2094916″]

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.[/hidden-answer]

Try It

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

[reveal-answer q=”fs-id2571919″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id2571919″]

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

[/hidden-answer]

Access these online resources for additional instruction and practice with sine and cosine functions.

• Trigonometric Functions Using the Unit Circle
• Sine and Cosine from the Unit
• Sine and Cosine from the Unit Circle and Multiples of Pi Divided by Six
• Sine and Cosine from the Unit Circle and Multiples of Pi Divided by Four
• Trigonometric Functions Using Reference Angles