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

$$\sin^2\frac{\pi}{12}=\frac{2-\sqrt3}{4}$$
$$\sin^2\frac{\pi}{12}$$ *Recall the half-angle formula for sine, which is $$\sin^2\theta=\frac{1-\cos2\theta}{2}$$ Thus, $$\sin^2\frac{\pi}{12}=\frac{1-\cos\frac{\pi}{6}}{2}$$ $$\sin^2\frac{\pi}{12}=\frac{1-\frac{\sqrt3}{2}}{2}$$ $$\sin^2\frac{\pi}{12}=\frac{\frac{2-\sqrt3}{2}}{2}$$ $$\sin^2\frac{\pi}{12}=\frac{2-\sqrt3}{4}$$