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: 50


$$\cos3x=4\cos^3x-3\cos x$$

Work Step by Step

$$\cos3x$$ The question asks to write $\sin4x$ in terms of a trigonometric function of $x$. We will need both Sum and Difference Identities and Double-Angle Identities to help here. $$\cos3x=\cos(2x+x)$$ - Sum Identity for cosine: $\cos(A+B)=\cos A\cos B-\sin A\sin B$ with here $A=2x$ and $B=x$ $$\cos3x=\cos2x\cos x-\sin2x\sin x$$ - For $\sin2x$, we apply Double-Angle Identity for sine: $\sin2A=2\sin A\cos A$ with $A=x$. - For $\cos2x$, we apply Double-Angle Identity for cosine: $\cos2A=\cos^2A-\sin^2A$. $$\cos3x=(\cos^2x-\sin^2x)\cos x-(2\sin x\cos x)\sin x$$ $$\cos3x=\cos^3x-\sin^2x\cos x-2\sin^2x\cos x$$ $$\cos3x=\cos^3x-3\sin^2x\cos x$$ - Finally, for $\sin^2x$, we apply Pythagorean Identity: $\sin^2x=1-\cos^2x$ $$\cos3x=\cos^3x-3(1-\cos^2x)\cos x$$ $$\cos3x=\cos^3x-3\cos x+3\cos^3x$$ $$\cos3x=4\cos^3x-3\cos x$$
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