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 - Summary Exercises on Verifying Trigonometric Identities - Page 245: 15


$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$ The proof is explained in the work step by step below.

Work Step by Step

$$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$ We examine the right side first. $$X=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$ - Half-angle identity for tangent: $$\tan\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{1+\cos x}}$$ So it follows that $$\tan^2\frac{x}{2}=\frac{1-\cos x}{1+\cos x}$$ Apply the identity to $X$ $$X=\frac{1-\frac{1-\cos x}{1+\cos x}}{1+\frac{1-\cos x}{1+\cos x}}$$ $$X=\frac{\frac{1+\cos x-1+\cos x}{1+\cos x}}{\frac{1+\cos x+1-\cos x}{1+\cos x}}$$ $$X=\frac{1+\cos x-1+\cos x}{1+\cos x+1-\cos x}$$ $$X=\frac{2\cos x}{2}$$ $$X=\cos x$$ Thus, $$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$ The equation has been verified to be 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.