22) $sin\frac{9pi}{4}$

Explanation:

Step1: Rewrite the angle

We know that $\frac{9\pi}{4}=2\pi+\frac{\pi}{4}$. According to the periodicity of the sine - function $\sin(x + 2k\pi)=\sin(x)$ where $k\in\mathbb{Z}$. Here $k = 1$ and $x=\frac{\pi}{4}$, so $\sin\frac{9\pi}{4}=\sin(2\pi+\frac{\pi}{4})$. Since $\sin(x + 2k\pi)=\sin(x)$, then $\sin(2\pi+\frac{\pi}{4})=\sin\frac{\pi}{4}$.

Step2: Recall the value of $\sin\frac{\pi}{4}$

We know that for the special - angle $\frac{\pi}{4}$ (or $45^{\circ}$), $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$.

Answer:

$\frac{\sqrt{2}}{2}$