Question

1. \\( \sin \frac { 7 x } { 6 } \\) a) \\( - \frac { 1 } { 2 } \\) b) \\( \frac { 1 } { 2 } \\) c) \\( - \frac { \sqrt { 3 } } { 2 } \\) d) \\( \frac { \sqrt { 3 } } { 2 } \\)

Explanation:

Step1: Convert angle to standard position

$\sin\frac{7\pi}{6}=\sin(\pi+\frac{\pi}{6})$

Step2: Use sine addition formula

According to $\sin(A + B)=\sin A\cos B+\cos A\sin B$, here $A = \pi$, $B=\frac{\pi}{6}$.
$\sin(\pi+\frac{\pi}{6})=\sin\pi\cos\frac{\pi}{6}+\cos\pi\sin\frac{\pi}{6}$
Since $\sin\pi = 0$, $\cos\pi=- 1$, $\sin\frac{\pi}{6}=\frac{1}{2}$, $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$
So $\sin(\pi+\frac{\pi}{6})=0\times\frac{\sqrt{3}}{2}+(-1)\times\frac{1}{2}=-\frac{1}{2}$

Answer:

A. $-\frac{1}{2}$