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: 34


See the explanation below.

Work Step by Step

$\cos 4t=8{{\cos }^{4}}t-8{{\cos }^{2}}t+1$ Recall the double angle formula, $\cos 2t={{\cos }^{2}}t-1$ Consider the left side of the given expression and apply the above formula, $\begin{align} & \cos 4t=\cos 2\left( 2t \right) \\ & =2{{\cos }^{2}}\left( 2t \right)-1 \\ & =2{{\left( 2{{\cos }^{2}}t-1 \right)}^{2}}-1 \end{align}$ Recall the algebraic square formula, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ Apply the algebraic square formula. $\begin{align} & \cos 4t=2\left( 4{{\cos }^{2}}t+1-4{{\cos }^{2}}t \right)-1 \\ & =\left( 8{{\cos }^{2}}t+2-8{{\cos }^{2}}t \right)-1 \\ & =8{{\cos }^{4}}t-8{{\cos }^{2}}t+1 \end{align}$ Hence, it is proved that the given identity $\cos 4t=8{{\cos }^{4}}t-8{{\cos }^{2}}t+1$ holds true.
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