Question

sin^6+cos^6=1-3sin^2cos^2

Answer 1

LHS = sin^6A + cos^6 A

= ( sin²A )³ + ( cos² A )³

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We know the algebraic identity :

x³ + y³ + 3xy( x + y ) = ( x + y )³

Or

x³ + y³ = ( x + y )³ - 3xy(x+y)

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Here , x = sin²A , y = cos²A

=(sin²A+cos²A)³-3sinAcosA(sin²A+cos²A)

= 1 - 3sinAcosA

[ Since , sin²A + cos²A = 1 ]

= RHS

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Answer 2

Solution :

Now, LHS = sin⁶θ + cos⁶θ

= (sin²θ)³ + (cos²θ)³

= (sin²θ + cos²θ) {(sin²θ)² + (cos²θ)² - sin²θ cos²θ}

= 1 * {(sin²θ + cos²θ)² - 2sin²θ cos²θ - sin²θ cos²θ}

= 1 - 3 sin²θ cos²θ

= RHS (Proved)