Question

use the unit circle to find the value of sin(7π/4) and periodic properties of trigonometric functions to find the value of sin(15π/4). answer boxes in your choice. a. sin(7π/4)=-√2/2 (type an exact answer, using radicals as needed. simplify your answer. rationalize the denominator.) b. the solution is undefined. select the correct choice below and fill in any answer boxes in your choice. a. sin(15π/4)= (type an exact answer, using radicals as needed. simplify your answer. rationalize the denominator.) b. the solution is undefined.

Explanation:

Step1: Analyze $\sin\frac{7\pi}{4}$

The angle $\frac{7\pi}{4}$ is in the fourth - quadrant. On the unit - circle, for an angle $\theta$, $\sin\theta$ is the $y$ - coordinate of the point on the unit - circle corresponding to the angle $\theta$. The angle $\frac{7\pi}{4}=2\pi-\frac{\pi}{4}$. The coordinates of the point on the unit - circle for $\theta = \frac{7\pi}{4}$ are $(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$, so $\sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}$.

Step2: Analyze $\sin\frac{15\pi}{4}$

We know that the period of the sine function is $2\pi$, i.e., $\sin(x + 2k\pi)=\sin(x)$ for any real number $x$ and integer $k$. We can rewrite $\frac{15\pi}{4}$ as $\frac{15\pi}{4}=4\pi-\frac{\pi}{4}$. Since $\sin(x + 4\pi)=\sin(x)$ (because $4\pi = 2\times2\pi$), then $\sin\frac{15\pi}{4}=\sin(4\pi-\frac{\pi}{4})$. Using the property $\sin(A - B)=\sin A\cos B-\cos A\sin B$ with $A = 4\pi$ and $B=\frac{\pi}{4}$, and knowing that $\sin(4\pi)=0$ and $\cos(4\pi)=1$, we get $\sin\frac{15\pi}{4}=-\frac{\sqrt{2}}{2}$.

Answer:

A. $\sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}$
A. $\sin\frac{15\pi}{4}=-\frac{\sqrt{2}}{2}$