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 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 30


$$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$$ The equation is an identity, as proved below.

Work Step by Step

$$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$$ From the right side, we examine first. $$X=\frac{1-\tan^2x}{1+\tan^2x}$$ - Quotient Identity: $\tan x=\frac{\sin x}{\cos x}$ Replace it into $X$: $$X=\frac{1-\Big(\frac{\sin x}{\cos x}\Big)^2}{1+\Big(\frac{\sin x}{\cos x}\Big)^2}$$ $$X=\frac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x}}$$ $$X=\frac{\frac{\cos^2x-\sin^2x}{\cos^2x}}{\frac{\cos^2x+\sin^2x}{\cos^2x}}$$ $$X=\frac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x}$$ - Double-Angle Identity: $\cos^2x-\sin^2x=\cos2x$ - Pythagorean Identity: $\cos^2x+\sin^2x=1$ Replace them into $X$: $$X=\frac{\cos2x}{1}$$ $$X=\cos2x$$ Therefore, $$\cos2x=\frac{1-\tan^2x}{1+\tan^2x}$$ 2 sides are equal, so this must 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.