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