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: 22

Answer

$$\sin2x=\frac{2\tan x}{1+\tan^2x}$$ The equation is an identity.

Work Step by Step

$$\sin2x=\frac{2\tan x}{1+\tan^2x}$$ The right side is more complex, so we need to deal with it first. $$X=\frac{2\tan x}{1+\tan^2x}$$ - From Quotient Identity: $$\tan x=\frac{\sin x}{\cos x}$$ Therefore, $$X=\frac{\frac{2\sin x}{\cos x}}{1+\frac{\sin^2x}{\cos^2x}}$$ $$X=\frac{\frac{2\sin x}{\cos x}}{\frac{\cos^2x+\sin^2x}{\cos^2x}}$$ $$X=\frac{(2\sin x)\cos^2x}{(\cos^2x+\sin^2x)\cos x}$$ $$X=\frac{2\sin x\cos x}{\cos^2x+\sin^2x}$$ - From Pythagorean Identity: $\cos^2x+\sin^2x=1$ - From Double-Angle Identity: $2\sin x\cos x=\sin2x$ $$X=\frac{\sin 2x}{1}$$ $$X=\sin2x$$ So, $$\sin2x=\frac{2\tan x}{1+\tan^2x}$$ The equation is an identity.
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