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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