Answers: 1
![\bf sin^2({{ \theta}})=\cfrac{1-cos(2{{ \theta}})}{2}\qquad \qquad cos^2({{ \theta}})=\cfrac{1+cos(2{{ \theta}})}{2}\\ -----------------------------\\ thus \\\\ 2sin^2(x)cos^2(x)\implies \boxed{2}\cdot \cfrac{1-cos(2{{ x}})}{\boxed{2}}\cdot \cfrac{1+cos(2{{ x}})}{2} \\\\ \cfrac{1-cos(2{{ x}})}{1}\cdot \cfrac{1+cos(2{{ x}})}{2}\implies \cfrac{[1-cos(2x)][1+cos(2x)]}{2}\\ -----------------------------\\ \textit{now recall}\implies \textit{difference of squares} \\ \quad \\ ](/tpl/images/0378/3541/db068.png)
![\bf (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\ -----------------------------\\ thus \\\\ \cfrac{[1-cos(2x)][1+cos(2x)]}{2}\implies \cfrac{1^2-cos^2(x)}{2} \\\\ \cfrac{1-cos^2(x)}{2}\impliedby \begin{array}{llll} \textit{pretty sure you know what that numerator is}\\ \textit{check your pythagorean identities} \end{array}](/tpl/images/0378/3541/e6409.png)
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