Answer
The exact value of the expression is $-1$.
Work Step by Step
Let us consider the following equation.
$y=g\left( \frac{5\pi }{6}+\frac{\pi }{6} \right)+g\left( \frac{5\pi }{6} \right)+g\left( \frac{\pi }{6} \right)$
The condition is:
$g\left( x \right)=\cos x$
Now, apply the condition in equation $y$.
$\begin{align}
& y=\cos \left( \frac{5\pi }{6}+\frac{\pi }{6} \right)+\cos \left( \frac{5\pi }{6} \right)+\cos \left( \frac{\pi }{6} \right) \\
& =\cos \left( \frac{5\pi +\pi }{6} \right)+\cos \left( \frac{5\pi }{6} \right)+\cos \left( \frac{\pi }{6} \right) \\
& =\cos \left( \frac{6\pi }{6} \right)+\cos \left( \frac{5\pi }{6} \right)+\cos \left( \frac{\pi }{6} \right)
\end{align}$
Solve further,
$y=\cos \left( \pi \right)+\cos \left( \frac{5\pi }{6} \right)+\cos \left( \frac{\pi }{6} \right)$ …… (1)
So, the angle $\frac{5\pi }{6}$ lies between $\frac{\pi }{2}$ and $\pi $, in quadrant II.
Calculate the reference angle as follows:
$\begin{align}
& {\theta }'=\pi -\frac{5\pi }{6} \\
& =\frac{6\pi -5\pi }{6} \\
& =\frac{\pi }{6}
\end{align}$
So,
$\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$
Here, the cosine is negative in quadrant II.
Thus,
$\cos \frac{5\pi }{6}=-\cos \frac{\pi }{6}$
Put $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$.
$\cos \frac{5\pi }{6}=-\frac{\sqrt{3}}{2}$
Put $-1$ for $\cos \pi $ , $-\frac{\sqrt{3}}{2}$ for $\cos \frac{5\pi }{6}$ and $\frac{\sqrt{3}}{2}$ for $\cos \left( \frac{\pi }{6} \right)$ in equation (1).
$\begin{align}
& y=\left( -1 \right)+\left( -\frac{\sqrt{3}}{2} \right)+\left( \frac{\sqrt{3}}{2} \right) \\
& =-1
\end{align}$
