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 5.1 Fundamental Identities - 5.1 Exercises - Page 201: 45



Work Step by Step

$$\cos\theta=\frac{x}{x+1}$$ According to the Pythagorean Identity $$\sin^2\theta+\cos^2\theta=1$$ we can rewrite as $$\sin^2\theta=1-\cos^2\theta$$ Then, $$\sin\theta=\sqrt{1-\cos^2\theta}$$ This is the way we would be able to find an expression of $x$ for $\sin\theta$ 1) Find $\cos^2\theta$ $$\cos^2\theta=\frac{x^2}{(x+1)^2}$$ 2) Find $\sin^2\theta$ $$\sin^2\theta=1-\cos^2\theta$$ $$\sin^2\theta=1-\frac{x^2}{(x+1)^2}$$ $$\sin^2\theta=\frac{(x+1)^2-x^2}{(x+1)^2}$$ $$\sin^2\theta=\frac{(x+1-x)(x+1+x)}{(x+1)^2}$$ (for $a^2-b^2=(a-b)(a+b)$) $$\sin^2\theta=\frac{1\times(2x+1)}{(x+1)^2}$$ $$\sin^2\theta=\frac{2x+1}{(x+1)^2}$$ 3) Find $\sin\theta$ $$\sin\theta=\sqrt{\sin^2\theta}$$ $$\sin\theta=\frac{\sqrt{2x+1}}{\sqrt{(x+1)^2}}$$ $$\sin\theta=\frac{\sqrt{2x+1}}{|x+1|}$$ (since $\sqrt{a^2}=|a|$) Unfortunately, we cannot eliminate the absolute value $||$ sign, since there is not enough information to know whether $(x+1)\gt0$ or $\lt0$.
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