## Precalculus (6th Edition) Blitzer

The exact value of the expression is $\sqrt{2}$.
Let us consider the following equation. $y=\left( h\circ f \right)\left( \frac{11\pi }{4} \right)$ The condition is \begin{align} & f\left( x \right)=\sin x \\ & h\left( x \right)=2x \\ \end{align} Now, apply the condition in equation $y$. $y=h\left( f\left( \frac{11\pi }{4} \right) \right)$ Further solve, $y=2\left( \sin \left( \frac{11\pi }{4} \right) \right)$ …… (1) The angle $\frac{11\pi }{4}$ lies between $\frac{5\pi }{2}$ and $3\pi$, in quadrant II. Calculate the reference angle as follows \begin{align} & {\theta }'=3\pi -\frac{11\pi }{4} \\ & =\frac{12\pi -11\pi }{4} \\ & =\frac{\pi }{4} \end{align} Hence, $\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ Here, the sine is positive in quadrant II. So, $\sin \frac{11\pi }{4}=\sin \frac{\pi }{4}$ Put $\frac{\sqrt{2}}{2}$ for $\sin \frac{\pi }{4}$. $\sin \frac{11\pi }{4}=\frac{\sqrt{2}}{2}$ Put $\frac{\sqrt{2}}{2}$ for $\sin \frac{11\pi }{4}$ in equation (1). \begin{align} & y=2\left( \frac{\sqrt{2}}{2} \right) \\ & =\sqrt{2} \end{align}