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No longer require the answer to this, but if you want to, feel free to try it so I can double check my math (but also my newest question is more urgent).

 

Prove this identity:

18sin2θcos2θ=cos(4θ)

 May 15, 2019
edited by AdamTaurus  May 15, 2019
edited by AdamTaurus  May 15, 2019
 #1
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Prove:

18sin2θcos2θ=cos(4θ)LHS=12(2sinθcosθ)2LHS=12sin2(2θ)LHS=1sin2(2θ)sin2(2θ)LHS=cos2(2θ)sin2(2θ)LHS=cos(4θ)LHS=RHSQED

 May 15, 2019

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