Question: Prove that sin4θ+cos4θ=1−2sin2θcos2θ
What I have attempted (Usually I start of with the complex side)
So starting with the LHS
sin4θ+cos4θ=1−2sin2θcos2θ
(sin2θ)2+cos4θ=1−2sin2θcos2θ
(1−cos2θ)2+cos4θ=1−2sin2θcos2θ
(1−cos2θ)(1−cos2θ)+cos4θ=1−2sin2θcos2θ
1−2cos2θ+cos4θ+cos4θ=1−2sin2θcos2θ
1−2cos2θ+2cos4θ=1−2sin2θcos2θ
Now I am stuck... is my approach correct?
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