## Trigonometry (11th Edition) Clone

$$\cos\frac{\pi}{12}=\frac{\sqrt2+\sqrt6}{4}$$
$$\cos\frac{\pi}{12}$$ In terms of radians, the angles $\frac{\pi}{6}$, $\frac{\pi}{4}$ and $\frac{\pi}{3}$ are already known. So we would try to rewrite $\frac{7\pi}{12}$ in terms of them. In the previous exercise, we see that rewriting $7\pi=4\pi+3\pi$ works, so here we would rewrite $\pi=4\pi-3\pi$. $$\frac{\pi}{12}=\frac{4\pi-3\pi}{12}=\frac{4\pi}{12}-\frac{3\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}$$ That means, $$\cos\frac{\pi}{12}=\cos(\frac{\pi}{3}-\frac{\pi}{4})$$ We then apply cosine of a difference: $$\cos\frac{\pi}{12}=\cos\frac{\pi}{3}\cos\frac{\pi}{4}+\sin\frac{\pi}{3}\sin\frac{\pi}{4}$$ $$\cos\frac{\pi}{12}=\frac{1}{2}\frac{\sqrt2}{2}+\frac{\sqrt3}{2}\frac{\sqrt2}{2}$$ $$\cos\frac{\pi}{12}=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}$$ $$\cos\frac{\pi}{12}=\frac{\sqrt2+\sqrt6}{4}$$