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24 February, 06:55

Use De Moivre's theorem to express cos 5θ and sin 5θ in terms of sin θ and cos θ.

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  1. 24 February, 08:21
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    From the de Moivre's we have,

    (cosθ+isinθ) ^n=cos (nθ) + isin (nθ)

    Therefore,

    R ((cosθ+isinθ) ^5) = cos (5θ) I ((cosθ+isinθ) ^5) = sin (5θ)

    Simplifying,

    cos^5 (θ) - 10 (sin^2 (θ)) (cos^3 (θ)) + 5 (sin^4 (θ)) (cosθ) = cos (5θ)
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