## University Calculus: Early Transcendentals (3rd Edition)

$$\cos\frac{\pi}{12}=\frac{\sqrt2+\sqrt6}{4}$$
$$\cos\frac{\pi}{12}$$ We see that: $$\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}$$ Therefore, $$\cos\frac{\pi}{12}=\cos\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$$ Here, remember the identity we have proved in Exercise 35: $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ Apply the identity here, we have: $$\cos\frac{\pi}{12}=\cos\frac{\pi}{3}\cos\frac{\pi}{4}+\sin\frac{\pi}{3}\sin\frac{\pi}{4}$$ $$\cos\frac{\pi}{12}=\frac{1}{2}\times\frac{\sqrt2}{2}+\frac{\sqrt3}{2}\times\frac{\sqrt2}{2}$$ $$\cos\frac{\pi}{12}=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}$$ $$\cos\frac{\pi}{12}=\frac{\sqrt2+\sqrt6}{4}$$