Tuesday, 8 April 2014

calculus - Trigonometry Identity (Proof): sin4theta+cos4theta=12sin2thetacos2theta



Question: Prove that sin4θ+cos4θ=12sin2θcos2θ






What I have attempted (Usually I start of with the complex side)



So starting with the LHS



sin4θ+cos4θ=12sin2θcos2θ



(sin2θ)2+cos4θ=12sin2θcos2θ



(1cos2θ)2+cos4θ=12sin2θcos2θ




(1cos2θ)(1cos2θ)+cos4θ=12sin2θcos2θ



12cos2θ+cos4θ+cos4θ=12sin2θcos2θ



12cos2θ+2cos4θ=12sin2θcos2θ



Now I am stuck... is my approach correct?

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