# Prove that 2(sin

Consider the left hand side:

2(sin 6 θ +cos 6 θ ) – 3(sin 4 θ + cos 4 θ ) + 1

= 2[(sin 2 θ ) 3 + (cos 2 θ ) 3 ] – 3[(sin 2 θ ) 2 + (cos 2 θ ) 2 ] + 1
= 2[(sin 2 θ + cos 2 θ ) 3 – 3sin 2 θ cos 2 θ (sin 2 θ + cos 2 θ )] – 3[(sin 2 θ + cos 2 θ ) 2 – 2sin 2 θ cos 2 θ ] + 1
[The algebraic identity a 3 + b 3 = (a + b) 3 – 3ab(a + b) and a 2 + b 2 = (a + b) 2 – 2ab
are used in the above step where a = sin 2 θ and b = cos 2 θ].

Since, sin 2 θ + cos 2 θ = 1, we have
L.H.S. = 2[ 1 – 3 sin 2 θ cos 2 θ ] – 3[1– 2 sin 2 θ cos 2 θ ] + 1
= 2 – 6 sin 2 θ cos 2 θ - 3 + 6 sin 2 θ cos 2 θ + 1
= - 3 + 3
= 0

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