Answer
The exact value of the expression is $\frac{\sqrt{6}-\sqrt{2}}{4}$.
Work Step by Step
Let us consider the following equation.
$x=\sin \frac{17\pi }{3}\cos \frac{5\pi }{4}+\cos \frac{17\pi }{3}\sin \frac{5\pi }{4}$ …… (1)
So, in quadrant I all the trigonometric functions are positive. In quadrant II sine and cosecant are positive and all other trigonometric functions are negative. In quadrant III tangent and cotangent are positive and all other trigonometric functions are negative. In quadrant IV cosine and secant are positive and all other trigonometric functions are negative.
The angle $\frac{17\pi }{3}$ lies between $\frac{11\pi }{2}$ and $6\pi $, in quadrant IV.
Now, 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,
$\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$
Here, the sine is negative in quadrant IV.
So,
$\sin \frac{17\pi }{3}=-\sin \frac{\pi }{3}$
Put $\frac{\sqrt{3}}{2}$ for $\sin \frac{\pi }{3}$.
$\sin \frac{17\pi }{3}=-\frac{\sqrt{3}}{2}$
The angle $\frac{5\pi }{4}$ lies between $\pi $ and $\frac{3\pi }{2}$ , in quadrant III.
Put, calculate the reference angle as follows
$\begin{align}
& {\theta }'=\frac{5\pi }{4}-\pi \\
& =\frac{5\pi -4\pi }{4} \\
& =\frac{\pi }{4}
\end{align}$
Hence,
$\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$
Here, the cosine is negative in quadrant III.
So,
$\cos \frac{5\pi }{4}=-\cos \frac{\pi }{4}$
Put $\frac{\sqrt{2}}{2}$ for $\cos \frac{\pi }{4}$.
$\cos \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}$
The angle $\frac{17\pi }{3}$ lies between $\frac{11\pi }{2}$ and $6\pi $ , in quadrant IV.
Therefore, 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 IV.
So,
$\cos \frac{17\pi }{3}=-\cos \frac{\pi }{3}$
Put $\frac{1}{2}$ for $\cos \frac{\pi }{3}$.
$\cos \frac{17\pi }{3}=\frac{1}{2}$
The angle $\frac{5\pi }{4}$ lies between $\pi $ and $\frac{3\pi }{2}$ , in quadrant III.
So, calculate the reference angle as follows
$\begin{align}
& {\theta }'=\frac{5\pi }{4}-\pi \\
& =\frac{5\pi -4\pi }{4} \\
& =\frac{\pi }{4}
\end{align}$
Hence,
$\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}$
Here, the sine is negative in quadrant III.
So,
$\sin \frac{5\pi }{4}=-\sin \frac{\pi }{4}$
Put $\frac{\sqrt{2}}{2}$ for $\sin \frac{\pi }{4}$.
$\sin \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}$
Put $-\frac{\sqrt{3}}{2}$ for $\sin \frac{17\pi }{3}$ , $-\frac{\sqrt{2}}{2}$ for $\cos \frac{5\pi }{4}$ , $\frac{1}{2}$ for $\cos \frac{17\pi }{3}$ and $-\frac{\sqrt{2}}{2}$ for $\sin \frac{5\pi }{4}$ in equation (1).
$\begin{align}
& x=\left( -\frac{\sqrt{3}}{2} \right)\left( -\frac{\sqrt{2}}{2} \right)+\left( \frac{1}{2} \right)\left( -\frac{\sqrt{2}}{2} \right) \\
& =\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\
& =\frac{\sqrt{6}-\sqrt{2}}{4}
\end{align}$
