## Precalculus (6th Edition) Blitzer

The exact value of the expression is: $-\left( \frac{1+\sqrt{3}}{2} \right)$.
Let us consider the following equation: $y=f\left( \frac{4\pi }{3}+\frac{\pi }{6} \right)+f\left( \frac{4\pi }{3} \right)+f\left( \frac{\pi }{6} \right)$ The condition is $f\left( x \right)=\sin x$ Apply the condition in equation $y$ as follows: \begin{align} & y=\sin \left( \frac{4\pi }{3}+\frac{\pi }{6} \right)+\sin \left( \frac{4\pi }{3} \right)+\sin \left( \frac{\pi }{6} \right) \\ & =\sin \left( \frac{8\pi +\pi }{6} \right)+\sin \left( \frac{4\pi }{3} \right)+\sin \left( \frac{\pi }{6} \right) \\ & =\sin \left( \frac{9\pi }{6} \right)+\sin \left( \frac{4\pi }{3} \right)+\sin \left( \frac{\pi }{6} \right) \end{align} Solve further as follows: $y=\sin \left( \frac{3\pi }{2} \right)+\sin \left( \frac{4\pi }{3} \right)+\sin \left( \frac{\pi }{6} \right)$ …… (1) The angle $\frac{4\pi }{3}$ lies between $\pi$ and $\frac{3\pi }{2}$, in the quadrant III. Calculate the reference angle as follows: \begin{align} & {\theta }'=\frac{4\pi }{3}-\pi \\ & =\frac{4\pi -3\pi }{3} \\ & =\frac{\pi }{3} \end{align} Therefore, $\sin \frac{4\pi }{3}=\frac{\sqrt{3}}{2}$ Here, the sine is negative in quadrant III. Therefore, $\sin \frac{4\pi }{3}=-\sin \frac{\pi }{3}$ Put $\frac{\sqrt{3}}{2}$ for $\sin \frac{\pi }{3}$ as follows: $\sin \frac{4\pi }{3}=-\frac{\sqrt{3}}{2}$ Put $-1$ for $\sin \left( \frac{3\pi }{2} \right)$ , $-\frac{\sqrt{3}}{2}$ for $\sin \frac{4\pi }{3}$ , and $\frac{1}{2}$ for $\sin \left( \frac{\pi }{6} \right)$ in equation (1) as follows: \begin{align} & y=\left( -1 \right)+\left( -\frac{\sqrt{3}}{2} \right)+\left( \frac{1}{2} \right) \\ & =\frac{-2-\sqrt{3}+1}{2} \\ & =\frac{-1-\sqrt{3}}{2} \\ & =-\left( \frac{1+\sqrt{3}}{2} \right) \end{align}