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 242: 21



Work Step by Step

$$\sin\theta=\frac{3}{5}\hspace{1.5cm}90^\circ\lt \theta\lt180^\circ\hspace{1.5cm}\tan\frac{\theta}{2}=?$$ Apply the half-angle identity for tangent $$\tan\frac{\theta}{2}=\frac{1-\cos \theta}{\sin\theta}$$ However, we don't have the value of $\cos\theta$ now. So we need to find $\cos\theta$. 1) Find $\cos\theta$ - Pythagorean Identities: $$\cos^2\theta=1-\sin^2\theta=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$$ $$\cos\theta=\pm\frac{4}{5}$$ $90^\circ\lt\theta\lt180^\circ$ denotes that the position of $\theta$ is in quadrant II, where cosines are negative. Thus, $\cos\theta\lt0$. $$\cos\theta=-\frac{4}{5}$$ 2) Find $\tan\frac{\theta}{2}$ Now we can find $\tan\frac{\theta}{2}$ according to the mentioned identity. $$\tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}$$ $$\tan\frac{\theta}{2}=\frac{1-\Big(-\frac{4}{5}\Big)}{\frac{3}{5}}$$ $$\tan\frac{\theta}{2}=\frac{1+\frac{4}{5}}{\frac{3}{5}}$$ $$\tan\frac{\theta}{2}=\frac{\frac{9}{5}}{\frac{3}{5}}$$ $$\tan\frac{\theta}{2}=\frac{9}{3}=3$$
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