## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 236: 9

#### Answer

$$\sin2x=\frac{4}{5}$$ $$\cos2x=-\frac{3}{5}$$

#### Work Step by Step

$$\tan x=2 \hspace{2cm}\cos x\gt0$$ $$\sin 2x=?\hspace{2cm}\cos2x=?$$ 1) First, we need to calculate $\sin x$ and $\cos x$ as they are essential to the calculations of $\sin 2x$ and $\cos 2x$. According to Pythagorean Identities: $$\tan^2x+1=\sec^2x$$ According to Reciprocal Identities: $$\sec x=\frac{1}{\cos x}$$ Therefore, $$\tan^2x+1=\frac{1}{\cos^2x}$$ $$\cos^2x=\frac{1}{\tan^2x+1}=\frac{1}{2^2+1}=\frac{1}{5}$$ $$\cos x=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}\hspace{1cm}(\cos x\gt0)$$ Now we find $\sin x$, using Quotient Identities: $$\tan x=\frac{\sin x}{\cos x}$$ $$\sin x=\tan x\cos x=2\times\frac{\sqrt5}{5}=\frac{2\sqrt5}{5}$$ 2) Now we can calculate $\sin2x$ and $\cos2x$ using Double-Angle Identities $$\sin2x=2\sin x\cos x=2\times\frac{2\sqrt5}{5}\times\frac{\sqrt5}{5}=\frac{4}{5}$$ $$\cos 2x=\cos^2x-\sin^2x=\Big(\frac{\sqrt5}{5}\Big)^2-\Big(\frac{2\sqrt5}{5}\Big)^2=\frac{1}{5}-\frac{4}{5}=-\frac{3}{5}$$

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