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Mathematics

Factorize: sin3 θ + cos3 θ

Hence, prove the following identity :

sin3θ+cos3θsin θ + cos θ\dfrac{\text{sin}^3 θ + \text{cos}^3 θ}{\text{sin θ + cos θ}} + sin θ cos θ = 1

Factorisation

ICSE 2024

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Answer

Factorizing,

⇒ sin3 θ + cos3 θ = (sin θ + cos θ)(sin2 θ + cos2 θ - sin θ cos θ)

⇒ sin3 θ + cos3 θ = (sin θ + cos θ)(1 - sin θ cos θ) ………..(1)

To prove,

sin3θ+cos3θsin θ + cos θ\dfrac{\text{sin}^3 θ + \text{cos}^3 θ}{\text{sin θ + cos θ}} + sin θ cos θ = 1

Substituting value of sin3 θ + cos3 θ from equation (1) in L.H.S. of above equation :

⇒ (sin θ + cos θ)(1 - sin θ cos θ)sin θ + cos θ+sin θ cos θ\dfrac{\text{(sin θ + cos θ)(1 - sin θ cos θ)}}{\text{sin θ + cos θ}} + \text{sin θ cos θ}

⇒ 1 - sin θ cos θ + sin θ cos θ

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that sin3θ+cos3θsin θ + cos θ+sin θ cos θ=1\dfrac{\text{sin}^3 θ + \text{cos}^3 θ}{\text{sin θ + cos θ}} + \text{sin θ cos θ} = 1.

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