# 7.3 Unit circle  (Page 2/11)

 Page 2 / 11

## Sine and cosine functions

If $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is a real number and a point $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ on the unit circle corresponds to a central angle $\text{\hspace{0.17em}}t,$ then

$\mathrm{cos}\text{\hspace{0.17em}}t=x$
$\mathrm{sin}\text{\hspace{0.17em}}t=y$

Given a point P $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ on the unit circle corresponding to an angle of $\text{\hspace{0.17em}}t,$ find the sine and cosine.

1. The sine of $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is equal to the y -coordinate of point
2. The cosine of $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is equal to the x -coordinate of point

## Finding function values for sine and cosine

Point $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ is a point on the unit circle corresponding to an angle of $\text{\hspace{0.17em}}t,$ as shown in [link] . Find $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{sin}\left(t\right).$

We know that $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the x -coordinate of the corresponding point on the unit circle and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the y -coordinate of the corresponding point on the unit circle. So:

$\begin{array}{ccc}\hfill x& =\mathrm{cos}\text{\hspace{0.17em}}t\hfill & =\frac{1}{2}\hfill \\ y& =\mathrm{sin}\text{\hspace{0.17em}}t\hfill & =\frac{\sqrt{3}}{2}\hfill \end{array}$

A certain angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ corresponds to a point on the unit circle at $\text{\hspace{0.17em}}\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\text{\hspace{0.17em}}$ as shown in [link] . Find $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t.$

$\mathrm{cos}\left(t\right)=-\frac{\sqrt{2}}{2},\mathrm{sin}\left(t\right)=\frac{\sqrt{2}}{2}$

## 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 $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y.$

## Calculating sines and cosines along an axis

Find $\text{\hspace{0.17em}}\text{cos}\left(90°\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{sin}\left(90°\right).$

Moving $\text{\hspace{0.17em}}90°\text{\hspace{0.17em}}$ counterclockwise around the unit circle from the positive x -axis brings us to the top of the circle, where the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates are $\text{\hspace{0.17em}}\left(0,1\right),$ as shown in [link] .

We can then use our definitions of cosine and sine.

The cosine of $\text{\hspace{0.17em}}90°\text{\hspace{0.17em}}$ is 0; the sine of $\text{\hspace{0.17em}}90°\text{\hspace{0.17em}}$ is 1.

Find cosine and sine of the angle $\text{\hspace{0.17em}}\pi .$

$\mathrm{cos}\left(\pi \right)=-1,\mathrm{sin}\left(\pi \right)=0$

## 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 $\text{\hspace{0.17em}}{x}^{2}+{y}^{2}=1.\text{\hspace{0.17em}}$ Because $\text{\hspace{0.17em}}x=\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}t,$ we can substitute for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ to get $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1.\text{\hspace{0.17em}}$ This equation, $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1,$ is known as the Pythagorean Identity    . See [link] .

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.

## Pythagorean identity

The Pythagorean Identity    states that, for any real number $\text{\hspace{0.17em}}t,$

${\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1$

Given the sine of some angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and its quadrant location, find the cosine of $\text{\hspace{0.17em}}t.$

1. Substitute the known value of $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ into the Pythagorean Identity.
2. Solve for $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t.$
3. Choose the solution with the appropriate sign for the x -values in the quadrant where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is located.

## Finding a cosine from a sine or a sine from a cosine

If $\text{\hspace{0.17em}}\mathrm{sin}\left(t\right)=\frac{3}{7}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is in the second quadrant, find $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right).$

If we drop a vertical line from the point on the unit circle corresponding to $\text{\hspace{0.17em}}t,$ we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem. See [link] .

Substituting the known value for sine into the Pythagorean Identity,

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

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

$\text{cos}\left(t\right)=-\frac{2\sqrt{10}}{7}$

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