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: 51


As proved before, the equation is an identity. $$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$

Work Step by Step

$$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$ The left side would be examined first. $$X=1-\tan^2\frac{\theta}{2}$$ - Half-angle identity for tangent: $\tan\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$ (we choose this one since here it involves $\tan^2\frac{\theta}{2}$) Thus, $$\tan^2\frac{\theta}{2}=\frac{1-\cos\theta}{1+\cos\theta}$$ (After squaring, the $\pm$ sign can be removed) Replace back to $X$: $$X=1-\frac{1-\cos\theta}{1+\cos\theta}$$ $$X=\frac{1+\cos\theta-(1-\cos\theta)}{1+\cos\theta}$$ $$X=\frac{1+\cos\theta-1+\cos\theta}{1+\cos\theta}$$ $$X=\frac{2\cos\theta}{1+\cos\theta}$$ Therefore, $$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$ The equation is an identity.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.