Prove that 2(sin 6 theta +cos 6 theta ) -3(sin 4 theta +cos 4 theta) +1 =0
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LHS=2(sin 6 θ+cos 6 θ)−3(sin 4 θ+cos 4 θ)+1
=2{(sin 2 θ+cos 2 θ) 3 −3sin 2 θcos 2 θ(sin 2 θ+cos 2 θ)}−3(sin 2 θ+cos 2 θ) 2 −2(sin 2 θcos 2 θ)}+1
We know, [sin²x+cos²x=1]
=2{1−3sin 2 θcos 2 θ}−3{1−2sin 2 θcos 2 θ}+1
=2−6sin 2 θcos 2 θ−3+6sin 2 θcos 2 θ+1
=0
Explanation:
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