My Math Forum Trig identities cos(x+3x)=sin4x/sin2x

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 March 1st, 2016, 02:57 PM #1 Newbie   Joined: Feb 2016 From: singapore Posts: 3 Thanks: 0 Trig identities cos(x+3x)=sin4x/sin2x Hey I found this problem, Can anyone solve this? Cos(x+3x)=sin4x/sin2x
 March 1st, 2016, 04:24 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,770 Thanks: 1424 Your equation is not an identity ... Left side is $\cos(4x)=2\cos^2(2x)-1$ Right side is $\dfrac{\sin(4x)}{\sin(2x)} =\dfrac{2\sin(2x) \cos(2x)}{\sin(2x)} = 2\cos(2x)$ results in a quadratic in $\cos(2x)$ ... $2\cos^2(2x)-2\cos(2x)-1=0$ $\cos(2x)=\dfrac{1 - \sqrt{3}}{2}$ This last equation yields four solutions of $x$ for $0 < x < 2\pi$ ... I'll leave that part for you to calculate. Last edited by skipjack; March 1st, 2016 at 09:56 PM.

 Tags 3xsin4x or sin2x, cosx, identities, trig

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