find the exact value. (a) sin(11π/6) (b) cos(11π/4) (c) tan(-2π/3) (d) csc(9π/2)

Explanation:

Step1: Rewrite $\frac{11\pi}{6}$

$\frac{11\pi}{6}=2\pi-\frac{\pi}{6}$. Using the identity $\sin(2\pi - \alpha)=-\sin\alpha$, we have $\sin(\frac{11\pi}{6})=\sin(2\pi-\frac{\pi}{6})=-\sin\frac{\pi}{6}$. Since $\sin\frac{\pi}{6}=\frac{1}{2}$, then $\sin(\frac{11\pi}{6})=-\frac{1}{2}$.

Step2: Rewrite $\frac{11\pi}{4}$

$\frac{11\pi}{4}=2\pi + \frac{3\pi}{4}$. Using the identity $\cos(2\pi+\alpha)=\cos\alpha$, we get $\cos(\frac{11\pi}{4})=\cos(2\pi + \frac{3\pi}{4})=\cos\frac{3\pi}{4}$. Since $\cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}$, then $\cos(\frac{11\pi}{4})=-\frac{\sqrt{2}}{2}$.

Step3: Use the property of tangent

$\tan(-\alpha)=-\tan\alpha$. So $\tan(-\frac{2\pi}{3})=-\tan\frac{2\pi}{3}$. And $\frac{2\pi}{3}=\pi-\frac{\pi}{3}$, $\tan(\pi - \alpha)=-\tan\alpha$, then $\tan\frac{2\pi}{3}=-\tan\frac{\pi}{3}=-\sqrt{3}$, so $\tan(-\frac{2\pi}{3})=\sqrt{3}$.

Step4: Rewrite $\frac{9\pi}{2}$

$\frac{9\pi}{2}=4\pi+\frac{\pi}{2}$. Using the identity $\csc(2k\pi+\alpha)=\csc\alpha$ ($k\in\mathbb{Z}$), we have $\csc(\frac{9\pi}{2})=\csc(4\pi+\frac{\pi}{2})=\csc\frac{\pi}{2}$. Since $\csc\frac{\pi}{2} = 1$, then $\csc(\frac{9\pi}{2})=1$.

Answer:

(a) $-\frac{1}{2}$ (b) $-\frac{\sqrt{2}}{2}$ (c) $\sqrt{3}$ (d) $1$