## Trigonometry (11th Edition) Clone

$$\cos\Big(-\frac{\pi}{12}\Big)=\frac{\sqrt2+\sqrt6}{4}$$
$$\cos\Big(-\frac{\pi}{12}\Big)=\cos\frac{-\pi}{12}$$ 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, $$\cos\Big(-\frac{\pi}{12}\Big)=\cos\frac{\pi}{12}=\frac{\sqrt2+\sqrt6}{4}$$