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

$$\cos^2\frac{5\pi}{12}=\frac{2-\sqrt3}{4}$$
$$\cos^2\frac{5\pi}{12}$$ *Recall the half-angle formula for cosine, which is $$\cos^2\theta=\frac{1+\cos2\theta}{2}$$ Thus, $$\cos^2\frac{5\pi}{12}=\frac{1+\cos\frac{5\pi}{6}}{2}$$ In the unit circle, angle $\frac{5\pi}{6}$ is the equivalent of angle $\frac{\pi}{6}$ but in the second quadrant, where cosine is negative. Therefore, $\cos\frac{5\pi}{6}=-\cos\frac{\pi}{6}=-\frac{\sqrt3}{2}$ So, return to the given function: $$\cos^2\frac{5\pi}{12}=\frac{1-\frac{\sqrt3}{2}}{2}$$ $$\cos^2\frac{5\pi}{12}=\frac{\frac{2-\sqrt3}{2}}{2}$$ $$\cos^2\frac{5\pi}{12}=\frac{2-\sqrt3}{4}$$