# Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 52

$1-8\sin^2x\cos^2x$

#### Work Step by Step

RECALL: (1) $\cos{(2A)}=\cos^2A−\sin^2A$ (2) $\sin{(2A)}=2\sin{A}\cos{A}$ With $4x=2(2x)$, using the double-angle identity for cosine above gives: \begin{align*} \cos{(4x)}&=\cos^2{(2x)}−\sin^2{(2x)}\\ &=\left(\cos{(2x)}\right)^2-\left(\sin{(2x)}\right)^2 \end{align*} Using the double-angle identities for cosine and sine gives: \begin{align*} \cos(4x)&=\left(\cos{(2x)}\right)^2−\left(\sin{(2x)}\right)^2\\ &=(\cos^2x−\sin^2x)^2−(2\sin{x}\cos{x})^2\\ &=\cos^4{x}-2\cos^2{x}\sin^2{x}+\sin^4{x}-4\sin^2x\cos^2x\\ &=\cos^4{x}+\sin^4{x}-6\sin2x\cos^2x\\ &=\cos^4{x}+\sin^4x + (2\sin^2x\cos^2x-8\sin^2x\cos^2x)\\ &=(\cos^4{x}+2\sin^2x\cos^2x\sin^4x)-8\sin^2x\cos^2x\\ &=(\cos^2x+\sin^2x)^2-8\sin^2x\cos^2x\\ &=1^2-8\sin^2x\cos^2x &\text{(use } sin^2x+\cos^2x=1)\\ &=1-8\sin^2x\cos^2x \end{align*}

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