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November 17th, 2017, 08:43 AM   #11
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$4\sin^2{x}\cos^2{x} = (2\sin{x}\cos{x})^2 = \sin^2(2x)$

$x = 22.5^\circ \implies \sin^2(2x) = \sin^2(45^\circ) = \left(\dfrac{\sqrt{2}}{2}\right)^2 = \dfrac{1}{2}$
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November 17th, 2017, 01:35 PM   #12
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Thank you! That makes much more sense to me.
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