Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 243: 60



Work Step by Step

$$\sin\frac{\theta}{2}=\frac{1}{m}$$ $$m=\frac{3}{2}$$ Replace $m=\frac{3}{2}$ into the formula. $$\sin\frac{\theta}{2}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$$ - From half-angle identity for sines: $$\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}$$ Therefore, $$\pm\sqrt{\frac{1-\cos\theta}{2}}=\frac{2}{3}$$ As $\sqrt{\frac{1-\cos\theta}{2}}\ge0$ for all $\theta$, the equation would happen when we pick the positive square root. $$\sqrt{\frac{1-\cos\theta}{2}}=\frac{2}{3}$$ $$\frac{1-\cos\theta}{2}=\frac{4}{9}$$ $$9(1-\cos\theta)=8$$ $$1-\cos\theta=\frac{8}{9}$$ $$\cos\theta=1-\frac{8}{9}=\frac{1}{9}$$ which means $$\theta\approx83.62^\circ$$
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