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 680: 36

Answer

The value of $10{{\cos }^{4}}x$ is $\frac{15}{4}+5\cos 2x+\frac{5}{4}\cos 4x$.

Work Step by Step

We have to find the value of $10{{\cos }^{4}}x$; various cos formulas are used which represent the answer in the form of cos. $\begin{align} & 10{{\cos }^{4}}x=10{{\cos }^{2}}x{{\cos }^{2}}x \\ & =10{{\left( \frac{1+\cos 2x}{2} \right)}^{2}} \\ & =10\left( \frac{1+2\cos 2x+{{\cos }^{2}}2x}{4} \right) \\ & =\frac{10+20\cos 2x+10{{\cos }^{2}}2x}{4} \end{align}$ And take the denominator of the equation with each different numerator. $\begin{align} & 10{{\cos }^{4}}x=\frac{10}{4}+\frac{20\cos 2x}{4}+\frac{10{{\cos }^{2}}2x}{4} \\ & =\frac{5}{2}+5\cos 2x+\frac{5}{4}\left( 1+\cos 4x \right) \\ & =\frac{5}{2}+5\cos 2x+\frac{5}{4}+\frac{5}{4}\cos 4x \\ & =\frac{15}{4}+5\cos 2x+\frac{5}{4}\cos 4x \end{align}$ So, in step 2 the identity of ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ is used and in the last step, $\frac{5}{4}$ is distributed and added into $\frac{5}{2}$. Thus, the value of $10{{\cos }^{4}}x$ is $\frac{15}{4}+5\cos 2x+\frac{5}{4}\cos 4x$.
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