5.2 Unit circle: sine and cosine functions  (Page 6/12)

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

   $$\cos (\mathrm{30°})=\frac{\sqrt{3}}{2}\text{and}\sin (\mathrm{30°})=\frac{1}{2}.$$

   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.

   $$\cos (\mathrm{150°})=-\frac{\sqrt{3}}{2}\text{and}\sin (\mathrm{150°})=\frac{1}{2}$$
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:
   $$\cos \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}\text{and}\sin \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}$$

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 $\pi .$

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

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.

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

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