Question

if a sin cube theta + b cos cube theta equal to sin theta cos theta and a sin theta - bcos theta=0 prove that a squared plus b squared equals to 1

Answer 1

thanks it will help you

Answer 2

a² + b²  = 1  if aSin³θ + bCos³θ = SinθCosθ

Step-by-step explanation:

aSin³θ + bCos³θ = SinθCosθ

aSinθ - bCosθ = 0

to be proved a² + b² = 1

aSinθ - bCosθ = 0

=> aSinθ = bCosθ

aSin³θ + bCos³θ = SinθCosθ

=> aSinθSin²θ + bCos³θ = SinθCosθ

=> bCosθSin²θ + bCos³θ = SinθCosθ

Dividing by Cosθ both sides

=> bSin²θ + bCos²θ = Sinθ

=> b(Sin²θ + Cos²θ) = Sinθ

=> b = Sinθ

aSinθ = bCosθ

=> aSinθ = SinθCosθ

=> a = Cosθ

a² + b² = Sin²θ + Cos²θ = 1

Hence proved that a² + b²  = 1  if aSin³θ + bCos³θ = SinθCosθ

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