Answer
The exact value of the expression is $-\frac{3\sqrt{3}}{2}$.
Work Step by Step
Consider the following equation.
$x=\sin \frac{3\pi }{2}\tan \left( -\frac{8\pi }{3} \right)+\cos \left( -\frac{5\pi }{6} \right)$ …… (1)
In the given expression, if there are angles greater than $2\pi $ or less than $-2\pi $, do the following: first, find the positive co-terminal angle of that angle. Then, find the reference angle of the co-terminal angle. Finally, evaluate the trigonometric function at the reference angle with the correct sign.
The measure of $\left( -\frac{8\pi }{3} \right)$ is less than $-2\pi $. Add $4\pi $ to $\left( -\frac{8\pi }{3} \right)$ , to get a positive co-terminal angle less than $2\pi $. Consider $\alpha $ to be the positive co-terminal angle.
$\begin{align}
& \alpha =\left( -\frac{8\pi }{3} \right)+4\pi \\
& =\frac{-8\pi +12\pi }{3} \\
& =\frac{4\pi }{3}
\end{align}$
We know that $\frac{4\pi }{3}$ lies in quadrant III. Subtract $\pi $ form $\frac{4\pi }{3}$, to find the reference angle. Consider ${\alpha }'$ to be the reference angle.
$\begin{align}
& {\alpha }'=\frac{4\pi }{3}-\pi \\
& =\frac{4\pi -3\pi }{3} \\
& =\frac{\pi }{3}
\end{align}$
Hence,
$\tan \frac{\pi }{3}=\sqrt{3}$
Tangent is positive in quadrant III.
Therefore,
$\tan \left( -\frac{8\pi }{3} \right)=\tan \frac{\pi }{3}$
Substitute $\sqrt{3}$ for $\tan \frac{\pi }{3}$.
$\tan \left( -\frac{8\pi }{3} \right)=\sqrt{3}$
Add $\pi $ to $\frac{-5\pi }{6}$, to find the reference angle.
$\begin{align}
& {{\alpha }'}'=\left( -\frac{5\pi }{6} \right)+\pi \\
& =\frac{-5\pi +6\pi }{6} \\
& =\frac{\pi }{6}
\end{align}$
Hence,
$\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$
Cosecant is negative in quadrant III.
Therefore,
$\cos \frac{-5\pi }{6}=-\cos \frac{\pi }{6}$
Substitute $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$.
$\cos \frac{-5\pi }{6}=-\frac{\sqrt{3}}{2}$
Substitute $\left( -1 \right)$ for $\sin \frac{3\pi }{2}$ , $\sqrt{3}$ for $\tan \left( -\frac{8\pi }{3} \right)$ and $\frac{\sqrt{3}}{2}$ for $\cos \frac{-5\pi }{6}$ in equation (1).
$\begin{align}
& x=\sin \frac{3\pi }{2}\tan \left( -\frac{8\pi }{3} \right)+\cos \left( -\frac{5\pi }{6} \right) \\
& =\left( -1 \right)\left( \sqrt{3} \right)+\left( -\frac{\sqrt{3}}{2} \right) \\
& =\frac{-2\sqrt{3}-\sqrt{3}}{2} \\
& =-\frac{3\sqrt{3}}{2}
\end{align}$
