Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 682: 92


To reduce the power of $\left( {{\cos }^{4}}x \right)$, first rewrite $\left( {{\cos }^{4}}x \right)$ as ${{\left( {{\cos }^{2}}x \right)}^{2}}$. Then expand $\left( {{\cos }^{2}}x \right)$ using power reducing formula. Replace the term $\left( {{\cos }^{2}}x \right)$ in expression ${{\left( {{\cos }^{2}}x \right)}^{2}}$ by the result obtained from power reducing formula and expand it. While expanding there will be another term $\left( {{\cos }^{2}}2x \right)$. Express this term also using power reducing formula. Final result is the power reduced form of ${{\cos }^{4}}x$.

Work Step by Step

Now, rewrite $\left( {{\cos }^{4}}x \right)$ as ${{\left( {{\cos }^{2}}x \right)}^{2}}$. Then expand $\left( {{\cos }^{2}}x \right)$ by using the power reducing formula: $\begin{align} & {{\cos }^{4}}x={{\left( {{\cos }^{2}}x \right)}^{2}} \\ & {{\cos }^{2}}x=\frac{1+\cos 2x}{2} \\ \end{align}$ So, $\begin{align} & {{\cos }^{4}}x={{\left( {{\cos }^{2}}x \right)}^{2}} \\ & ={{\left( \frac{1+\cos 2x}{2} \right)}^{2}} \\ & =\frac{1}{4}{{\left( 1+\cos 2x \right)}^{2}} \end{align}$ Expand ${{\left( 1+\cos 2x \right)}^{2}}$ using ${{\left( a+b \right)}^{2}}.$ This means: ${{\left( 1+\cos 2x \right)}^{2}}=1+2\cos 2x+{{\cos }^{2}}2x$ Now, replace $x\text{ by }2x$ in $\left( {{\cos }^{2}}x=\frac{1+\cos 2x}{2} \right)$ to get $\left( {{\cos }^{2}}2x \right)$ That means: $\begin{align} & {{\cos }^{2}}2x=\frac{1+\cos 2.2x}{2} \\ & =\frac{1+\cos 4x}{2} \end{align}$ Use ${{\cos }^{2}}2x=\frac{1+\cos 4x}{2}$ to expand $\left( 1+2\cos 2x+{{\cos }^{2}}2x \right).$ That is: $\begin{align} & 1+2\cos 2x+{{\cos }^{2}}2x=1+2\cos 2x+\frac{1+\cos 4x}{2} \\ & =\frac{2+4\cos 2x+1+\cos 4x}{2} \\ & =\frac{3+4\cos 2x+\cos 4x}{2} \\ & {{\left( 1+\cos 2x \right)}^{2}}=\frac{3+4\cos 2x+\cos 4x}{2} \end{align}$ Thus, $\begin{align} & {{\cos }^{4}}x=\frac{1}{4}{{\left( 1+\cos 2x \right)}^{2}} \\ & =\frac{3+4\cos 2x+\cos 4x}{4.2} \\ & =\frac{3+4\cos 2x+\cos 4x}{8} \end{align}$
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