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November 17th, 2017, 07:43 AM  #11 
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 
$4\sin^2{x}\cos^2{x} = (2\sin{x}\cos{x})^2 = \sin^2(2x)$ $x = 22.5^\circ \implies \sin^2(2x) = \sin^2(45^\circ) = \left(\dfrac{\sqrt{2}}{2}\right)^2 = \dfrac{1}{2}$ 
November 17th, 2017, 12:35 PM  #12 
Newbie Joined: Nov 2017 From: Pennsylvania Posts: 7 Thanks: 0 
Thank you! That makes much more sense to me.


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identities, solving or verifying, trig 
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