Use the addition or subtraction formula for cosine to compute $\cos(-5\pi/12)$ Use the addition or subtraction formula for cosine to compute $\cos(-5\pi/12)$ (Leave your answer in exact form.)
I have $$\begin{align}\cos(-5\pi/12)&=\cos((\pi/4)-(5\pi/6))\\
                   &=\cos(\pi/4)\cos(5\pi/6)+\sin(\pi/4)\sin(5\pi/6)\\
                   &=\frac{\sqrt2}2\cdot\frac{\sqrt3}2+\frac{\sqrt2}2\cdot\frac 12\end{align}$$
Is this right?
 A: Alternative, you can use the half-angle formula (I am aware that this was not specified):
$$\cos{\left(-\frac{5 \pi}{12}\right)} = \sqrt{\frac{1+\cos{(-5 \pi/6)}}{2}} = \frac{\sqrt{2-\sqrt{3}}}{2}$$
Now, $2-\sqrt{3} = (\sqrt{3}-1)^2/2$
so that
$$\cos{\left(-\frac{5 \pi}{12}\right)} = \frac{\sqrt{3}-1}{2 \sqrt{2}} = \frac{\sqrt{6}-\sqrt{2}}{4}$$
A: Hint: $\frac{5}{12}=\frac{1}{4}+\frac{1}{6}$. 
Alternately, $-\frac{5}{12}=\frac{1}{4}-\frac{2}{3}$.
There are other choices.
A: Maybe rewriting $-\dfrac{5\pi}{12}$ radians as $-75^\circ$ would help to immediately see that $-45^\circ-30^\circ = -75^\circ$ or that the half of $-150^\circ$ is $-75^\circ$.
A: Hint: use the identity $\cos(\alpha+\beta)\equiv\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$.
Letting $\alpha=\frac{\pi}{6}, \beta=\frac{\pi}{4}$, we get:
$\cos(\frac{5\pi}{12})=\cos(\frac{\pi}{6}+\frac{\pi}{4})=\cos(\frac{\pi}{6})\cos(\frac{\pi}4{})-\sin(\frac{\pi}{6})\sin(\frac{\pi}{4})=\frac{\sqrt{3}}{2}\cdot \frac{1}{\sqrt{2}}-\frac{1}{2} \cdot \frac{1}{\sqrt{2}}=\frac{1}{2\sqrt{2}}[\sqrt{3}-1]=\frac{\sqrt{3}-1}{2\sqrt{2}}=\cos(\frac{5\pi}{12})$
Now, us the identity $\cos(-\gamma)\equiv\cos(\gamma)$ (for $\gamma=\frac{5\pi}{12}$) to evaluate $\cos(\frac{-5\pi}{12}).$
