## Precalculus (6th Edition) Blitzer

The exact value of the expression is $-\frac{3}{2}$.
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}