Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.4 - Product-to-Sum and Sum-to-Product Formulas - Exercise Set - Page 691: 61


See the explanation below.

Work Step by Step

Let us consider the right-hand side of the provided expression: $\sin 2x+\sin 4x+\sin 6x$ Then, rearranging the terms: $\sin 4x+\left( \sin 2x+\sin 6x \right)$ One of the sum-to-product formulas is $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$. Therefore, $\sin 2x+\sin 6x$ can be written as provided below: $\sin 2x+\sin 6x=2\sin \frac{2x+6x}{2}\cos \frac{2x-6x}{2}$ Then, the expression can be computed as shown below: $\begin{align} & \sin 4x+\left( \sin 2x+\sin 6x \right)=\sin 4x+2\sin \left( \frac{2x+6x}{2} \right)\cos \left( \frac{2x-6x}{2} \right) \\ & =\sin 4x+2\sin \left( \frac{8x}{2} \right)\cos \left( \frac{-4x}{2} \right) \\ & =\sin 4x+2\sin 4x\cos \left( -2x \right) \end{align}$ By applying the even-odd identity, which is $cos(-x)=\cos x$, the expression can be further evaluated as given below: $\begin{align} & \sin 4x+2\sin 4x\cos \left( -2x \right)=\sin 4x+2\sin 4x\cos 2x \\ & =\sin \left( 2\cdot 2x \right)+2\sin 4x\cos 2x \end{align}$ Also, apply the double angle formula, which is $\sin 2\theta =2\sin \theta \cos \theta $, so, the expression can be further solved as: $\sin \left( 2\cdot 2x \right)+2\sin 4x\cos 2x=2\sin 2x\cos 2x+2\sin 4x\cos 2x$ By taking out the common terms: $2\sin 2x\cos 2x+2\sin 4x\cos 2x=2\cos 2x\left( \sin 2x+\sin 4x \right)$ Again, applying the sum-to-product formula, which is $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$, the expression can be calculated as shown below: $\begin{align} & 2\cos 2x\left( \sin 2x+\sin 4x \right)=2\cos 2x\left[ 2\sin \left( \frac{2x+4x}{2} \right)\cos \left( \frac{2x-4x}{2} \right) \right] \\ & =2\cos 2x\cdot 2\sin \left( \frac{6x}{2} \right)\cos \left( \frac{-2x}{2} \right) \\ & =2\cos 2x\cdot 2\sin 3x\cos \left( -x \right) \end{align}$ Then, apply the even-odd identity, which is $cos(-x)=\cos x$: $\begin{align} & 2\cos 2x\cdot 2\sin 3x\cos \left( -x \right)=2\cos 2x\cdot 2\sin 3x\cos x \\ & =4\cos 2x\sin 3x\cos x \end{align}$ Then rearrange the terms: $4\cos 2x\sin 3x\cos x=4\cos x\cos 2x\sin 3x$ Thus, the right-hand side of the expression is equal to the left-hand side: $4\cos x\cos 2x\sin 3x=\sin 2x+\sin 4x+\sin 6x$.
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