<s>[INST] Use an appropriate Half-Angle Formula to find the exact value of the expression $\cos \left(\frac{9\pi}{ 8}\right)$ I've been having problems all semester understanding radian fractions. If I were to double this fraction wouldn't it be $\frac{18\pi}{8}$. Yet, how does that help me find a reference angle on the unit circle? What is the best way for me to find a reference angle? I am familiar with $\frac{\pi}6$, $\frac{\pi}4$, $\frac{\pi}3$ but I am lost with radians that I can not convert to degrees.
Can someone explain. I have no problems with degrees but when it comes to radians that I am unable to convert to degrees, I am stuck.
 [/INST] From Gerry Myerson's comment, we have 
\begin{align*}
\frac{18\pi}{8} &= \frac{9\pi}{4} \\
&=2\pi+\frac{\pi}{4} \\
\cos \frac{18\pi}{8} &= \cos \left(2\pi+\frac{\pi}{4}\right) \\
&= \cos \frac{\pi}{4} \\
&= \frac{\sqrt{2}}{2}
\end{align*}
We then have the half-angle formula
$$
\cos \frac{1}{2}\theta = (-1)^{\lfloor (\theta + \pi) / (2\pi) \rfloor} \sqrt{\frac{1 + \cos \theta}{2}}
$$
Setting $\theta = \frac{18\pi}{8}$, we have
\begin{align*}
\cos \left(\frac{1}{2}\right)\left(\frac{18\pi}{8}\right) &= (-1)^{\lfloor ((18\pi/8) + \pi) / (2\pi) \rfloor} \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \\
\cos \frac{9\pi}{8} &= (-1)^{\lfloor(13\pi/4)/(2\pi)\rfloor}\sqrt{\frac{2 + \sqrt{2}}{4}} \\
&= (-1)^{\lfloor 13\pi / 8\rfloor} \frac{\sqrt{2 + \sqrt{2}}}{2} \\
&= (-1)^1 \frac{\sqrt{2 + \sqrt{2}}}{2} \\
\cos \frac{9\pi}{8} &= - \frac{\sqrt{2 + \sqrt{2}}}{2}
\end{align*}
</s>