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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