## Precalculus: Mathematics for Calculus, 7th Edition

a. $\sin 15^\circ$ b. $\sin 4\theta$
a. Use the identity $\sin \frac{u}{2}=\pm\sqrt{\frac{1-\cos u}{2}}$. Then $\sqrt{\frac{1-\cos 30^\circ}{2}}=\sin\frac{30^\circ}{2}=\sin 15^\circ$, and we can take the positive square root since $15^\circ$ is in Quadrant I, where sine is positive. b. Use the identity $\sin \frac{u}{2}=\pm\sqrt{\frac{1-\cos u}{2}}$. Then $\sqrt{\frac{1-\cos 8\theta}{2}}=\sin\frac{8\theta}{2}=\sin 4\theta$, with the restriction that $4\theta$ is in Quadrant I or Quadrant II, where sine is positive.