Answer
The exact value of the expression is $-\frac{3}{2}$.
Work Step by Step
Consider the following equation.
$x=\sin \frac{3\pi }{2}\tan \left( -\frac{15\pi }{4} \right)-\cos \left( -\frac{5\pi }{3} \right)$ …… (1)
In the given expression, if there are angles greater than $2\pi $ or less than $-2\pi $, 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 $\left( -\frac{15\pi }{4} \right)$ is less than $-2\pi $. Add $4\pi $ to $\left( -\frac{15\pi }{4} \right)$, to get a positive co-terminal angle less than $2\pi $.
$\begin{align}
& \alpha =\left( -\frac{15\pi }{4} \right)+4\pi \\
& =\frac{-15\pi +16\pi }{4} \\
& =\frac{\pi }{4}
\end{align}$
Hence,
$\tan \frac{\pi }{4}=1$
Tangent is positive in quadrant I.
Therefore,
$\tan \left( -\frac{15\pi }{4} \right)=\tan \frac{\pi }{4}$
Substitute $1$ for $\tan \frac{\pi }{4}$.
$\tan \left( -\frac{15\pi }{4} \right)=1$
Add $2\pi $ to $\frac{-5\pi }{3}$, to find the reference angle.
$\begin{align}
& {\alpha }'=\left( -\frac{5\pi }{3} \right)+2\pi \\
& =\frac{-5\pi +6\pi }{3} \\
& =\frac{\pi }{3}
\end{align}$
Hence,
$\cos \frac{\pi }{3}=\frac{1}{2}$
Cosecant is positive in quadrant I.
Therefore,
$\cos \frac{-5\pi }{3}=\cos \frac{\pi }{3}$
Substitute $\frac{1}{2}$ for $\cos \frac{\pi }{3}$
$\cos \frac{-5\pi }{3}=\frac{1}{2}$
Substitute $\left( -1 \right)$ for $\sin \frac{3\pi }{2}$ , $1$ for $\tan \left( -\frac{15\pi }{4} \right)$ and $\frac{1}{2}$ for $\cos \frac{-5\pi }{3}$ in equation (1).
$\begin{align}
& x=\sin \frac{3\pi }{2}\tan \left( -\frac{15\pi }{4} \right)-\cos \left( -\frac{5\pi }{4} \right) \\
& =\left( -1 \right)\left( 1 \right)-\left( \frac{1}{2} \right) \\
& =\frac{-2-1}{2} \\
& =-\frac{3}{2}
\end{align}$
