Answer
The exact value of the expression is: $-\left( \frac{1+\sqrt{3}}{2} \right)$.
Work Step by Step
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:
$\begin{align}
& y=\left( -1 \right)+\left( -\frac{\sqrt{3}}{2} \right)+\left( \frac{1}{2} \right) \\
& =\frac{-2-\sqrt{3}+1}{2} \\
& =\frac{-1-\sqrt{3}}{2} \\
& =-\left( \frac{1+\sqrt{3}}{2} \right)
\end{align}$
