## Precalculus (6th Edition) Blitzer

The exact value of the expression is $-1$.
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}