Q.288. Trigo identities.

Question 288.
Use De Moivre's theorem to derive identities for cos3θ  and sin3θ  in terms of cosθ and sinθ.

Answer 288.
(cosθ + isinθ)^3 = cos3θ + isin3θ

LHS
= cos^3 θ + 3(cos^2 θ)(isinθ) + 3cosθ(isinθ)^2 + (isinθ)^3
= cos^3 θ + (3cos^2 θ * sinθ) i - 3cosθ sin^2 θ - (sin^3 θ) i
=> cos3θ + isin3θ = cos^3 θ + (3cos^2 θ * sinθ) i - 3cosθ sin^2 θ - (sin^3 θ) i

Comparing real and imaginary parts,
cos3θ
= cos^3 θ - 3cosθ sin^2 θ
= cos^3 θ - 3cosθ (1 - cos^2 θ)
= 4cos^3 θ - 3cosθ
and
sin3θ
= 3cos^2 θ * sinθ - sin^3 θ
= 3(1 - sin^2 θ) * sinθ - sin^3 θ
= 3sinθ - 4sin^3 θ.

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