Prove that 2(sin 6 theta +cos 6 theta ) -(sin 4 theta +cos 4 theta) +1 =0
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2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1
=2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1
=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2
-2sin2θcos2θ]+1
The algebraic identity
a3 + b3 = (a+b)3 - 3ab(a+b) and
a2 + b2 = (a+b)2 - 2ab
are used in the above step where
a = sin2θ and b = cos2θ.
writing sin2θ + cos2θ = 1, we have
= 2[1-3 sin2θ cos2θ] -3[-2 sin2θcos2θ] + 1
= 2-6 sin2θcos2θ -3 + 6 sin2θcos2θ + 1
= -3+3=0
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