Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 230: 20

Answer

$$(\cos2x-\sin2x)^2=1-\sin4x$$ Since 2 sides are equal as proved above, the equation is an identity.

Work Step by Step

$$(\cos2x-\sin2x)^2=1-\sin4x$$ The left side would be tackled first here. $$X=(\cos2x-\sin2x)^2$$ Apply the expansion $(a-b)^2=a^2-2ab+b^2$ for $a=\cos2x$ and $b=\sin2x$. $$X=\cos^22x-2\cos2x\sin2x+\sin^22x$$ $$X=(\cos^22x+\sin^22x)-(2\cos2x\sin2x)$$ - From Pythagorean Identity: $\cos^22x+\sin^22x=1$ - And from Double-Angle Identity: $2\cos2x\sin2x=\sin(2\times2x)=\sin4x$ (for $\sin2\theta=2\sin\theta\cos\theta$) $$X=1-\sin4x$$ Therefore, $$(\cos2x-\sin2x)^2=1-\sin4x$$ Since 2 sides are equal, 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.