given f(x) = sin x, evaluate the function at x = 13π/2. (1 point)\n0\n-1\n1/2\n1

Explanation:

Step1: Rewrite the angle

We can rewrite $\frac{13\pi}{2}$ as $6\pi+\frac{\pi}{2}$. Since the sine - function has a period of $2\pi$, $\sin(x + 2k\pi)=\sin(x)$ for any real number $x$ and integer $k$. Here $x=\frac{\pi}{2}$ and $k = 3$. So, $\sin(\frac{13\pi}{2})=\sin(6\pi+\frac{\pi}{2})$.

Step2: Apply the periodicity property

Using the periodicity property of the sine - function $\sin(6\pi+\frac{\pi}{2})=\sin(\frac{\pi}{2})$.

Step3: Evaluate $\sin(\frac{\pi}{2})$

We know that $\sin(\frac{\pi}{2}) = 1$.

Answer:

1