Answer
$$\cos\Big(-\frac{\pi}{12}\Big)=\frac{\sqrt2+\sqrt6}{4}$$
Work Step by Step
$$\cos\Big(-\frac{\pi}{12}\Big)=\cos\frac{-\pi}{12}$$
In terms of radians, we would always try to rewrite the angles in terms of 3 angles: $\frac{\pi}{6}$, $\frac{\pi}{4}$ and $\frac{\pi}{3}$.
For $\pi$, we have $\pi=4\pi-3\pi$. So for $-\pi$, we have $-\pi=3\pi-4\pi$
$$\frac{-\pi}{12}=\frac{3\pi-4\pi}{12}=\frac{3\pi}{12}-\frac{4\pi}{12}=\frac{\pi}{4}-\frac{\pi}{3}$$
That means,
$$\cos\Big(-\frac{\pi}{12}\Big)=\cos\Big(\frac{\pi}{4}-\frac{\pi}{3}\Big)$$
We then apply cosine of a difference:
$$\cos\Big(-\frac{\pi}{12}\Big)=\cos\frac{\pi}{4}\cos\frac{\pi}{3}+\sin\frac{\pi}{4}\sin\frac{\pi}{3}$$
$$\cos\Big(-\frac{\pi}{12}\Big)=\frac{\sqrt2}{2}\frac{1}{2}+\frac{\sqrt2}{2}\frac{\sqrt3}{2}$$
$$\cos\Big(-\frac{\pi}{12}\Big)=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}$$
$$\cos\Big(-\frac{\pi}{12}\Big)=\frac{\sqrt2+\sqrt6}{4}$$
*Another way if you have done Exercise 12:
As we all know, $\cos(-\theta)=\cos\theta$.
Therefore, $$\cos\Big(-\frac{\pi}{12}\Big)=\cos\frac{\pi}{12}=\frac{\sqrt2+\sqrt6}{4}$$
