Answer
The exact value of the expression is $1$.
Work Step by Step
Let us consider the following equation.
$y=\left( h\circ g \right)\left( \frac{17\pi }{3} \right)$
And the condition is
$\begin{align}
& g\left( x \right)=\cos x \\
& h\left( x \right)=2x \\
\end{align}$
Apply the condition in equation $y$.
$y=h\left( g\left( \frac{17\pi }{3} \right) \right)$
Further solve,
$y=2\left( \cos \left( \frac{17\pi }{3} \right) \right)$ …… (1)
The angle $\frac{17\pi }{3}$ lies between $6\pi $ and $\frac{3\pi }{2}$, in quadrant III.
And, calculate the reference angle as follows
$\begin{align}
& {\theta }'=6\pi -\frac{17\pi }{3} \\
& =\frac{18\pi -17\pi }{3} \\
& =\frac{\pi }{3}
\end{align}$
Hence,
$\cos \frac{\pi }{3}=\frac{1}{2}$
Here, the cosine is positive in quadrant III.
So,
$\cos \frac{17\pi }{3}=\cos \frac{\pi }{3}$
Put $\frac{1}{2}$ for $\cos \frac{\pi }{3}$.
$\cos \frac{17\pi }{3}=\frac{1}{2}$
Put $\frac{1}{2}$ for $\cos \left( \frac{17\pi }{3} \right)$ in equation (1).
$\begin{align}
& y=2\left( \frac{1}{2} \right) \\
& =1
\end{align}$
