## Trigonometry (11th Edition) Clone

$$\cos\frac{7\pi}{12}=\frac{\sqrt2-\sqrt6}{4}$$
$$\cos\frac{7\pi}{12}$$ 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. $$\frac{7\pi}{12}=\frac{3\pi+4\pi}{12}=\frac{3\pi}{12}+\frac{4\pi}{12}=\frac{\pi}{4}+\frac{\pi}{3}$$ That means, $$\cos\frac{7\pi}{12}=\cos(\frac{\pi}{4}+\frac{\pi}{3})$$ We apply cosine of a sum: $$\cos\frac{7\pi}{12}=\cos\frac{\pi}{4}\cos\frac{\pi}{3}-\sin\frac{\pi}{4}\sin\frac{\pi}{3}$$ $$\cos\frac{7\pi}{12}=\frac{\sqrt2}{2}\frac{1}{2}-\frac{\sqrt2}{2}\frac{\sqrt3}{2}$$ $$\cos\frac{7\pi}{12}=\frac{\sqrt2}{4}-\frac{\sqrt6}{4}$$ $$\cos\frac{7\pi}{12}=\frac{\sqrt2-\sqrt6}{4}$$