## Trigonometry (11th Edition) Clone

$$\cos\Big(-\frac{7\pi}{12}\Big)=\frac{\sqrt2-\sqrt6}{4}$$
$$\cos\Big(-\frac{7\pi}{12}\Big)=\cos\frac{-7\pi}{12}$$ 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 $7\pi$, we have $\pi=4\pi+3\pi$. So for $-7\pi$, we have $-7\pi=-4\pi-3\pi$ $$\frac{-7\pi}{12}=\frac{-4\pi-3\pi}{12}=\frac{-4\pi}{12}-\frac{3\pi}{12}=-\frac{\pi}{3}-\frac{\pi}{4}$$ That means, $$\cos\Big(-\frac{7\pi}{12}\Big)=\cos\Big(-\frac{\pi}{3}-\frac{\pi}{4}\Big)$$ We then apply cosine of a difference: $$\cos\Big(-\frac{7\pi}{12}\Big)=\cos\Big(-\frac{\pi}{3}\Big)\cos\frac{\pi}{4}+\sin\Big(-\frac{\pi}{3}\Big)\sin\frac{\pi}{4}$$ As $\cos(-\theta)=\cos\theta$ and $\sin(-\theta)=-\sin\theta$, that makes $$\cos\Big(-\frac{7\pi}{12}\Big)=\cos\frac{\pi}{3}\cos\frac{\pi}{4}-\sin\frac{\pi}{3}\sin\frac{\pi}{4}$$ $$\cos\Big(-\frac{7\pi}{12}\Big)=\frac{1}{2}\frac{\sqrt2}{2}-\frac{\sqrt3}{2}\frac{\sqrt2}{2}$$ $$\cos\Big(-\frac{7\pi}{12}\Big)=\frac{\sqrt2}{4}-\frac{\sqrt6}{4}$$ $$\cos\Big(-\frac{7\pi}{12}\Big)=\frac{\sqrt2-\sqrt6}{4}$$ *Again, another way if you have done Exercise 11: As we all know, $\cos(-\theta)=\cos\theta$. Therefore, $$\cos\Big(-\frac{7\pi}{12}\Big)=\cos\frac{7\pi}{12}=\frac{\sqrt2-\sqrt6}{4}$$