Q) prove that sin 6 θ + cos 6 θ = 1 – 3 sin 2 θ cos 2 θ
Ans: Let’s start from LHS
LHS = prove that sin 6 θ + cos 6 θ
= (sin 2 θ) 3 + (cos 2 θ) 3
We know that a 3 + b 3 = (a + b) (a 2 + b 2 – a b)
∴ LHS = (sin 2 θ) 3 + (cos 2 θ) 3
= (sin 2 θ + cos 2 θ) ((sin 2 θ)2 + (cos 2 θ)2 – sin 2 θ cos 2 θ)
= ((sin 2 θ)2 + (cos 2 θ) 2 – sin 2 θ cos 2 θ) [∵ sin 2 θ + cos 2 θ = 1]
We know that, (a + b)2 = (a 2 + b 2 + 2 a b)
∴ a 2 + b 2 = (a + b)2 – 2 a b
∴ LHS = ((sin 2 θ)2 + (cos 2 θ) 2 – sin 2 θ cos 2 θ)
= ((sin 2 θ + cos 2 θ) 2 – 2 sin 2 θ cos 2 θ – sin 2 θ cos 2 θ)
= 1 – 3 sin 2 θ cos 2 θ
= RHS
Hence Proved !
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