Q.352. Trigonometric identities.

Question 352.
1) Expand sin^3θ cos^4θ in terms of sines of multiples of θ.
2) Prove that - 64sin^7x = sin7x - 7sin5x + 21sin3x - 35sinx.

Answer 352.
1)
sin^3 θ cos^4 θ
= (1/8) (2sinθ cosθ)^3 cosθ
= (1/8) sin^3 (2θ) cosθ
= (1/32) [3sin(2θ) - sin(6θ)] cosθ
= (1/32) [3sin(2θ) cosθ - sin(6θ) cosθ]
= (1/64) [3 (sin3θ + sinθ) - (sin7θ + sin5θ)]
= (1/64) (3sinθ + 3sin3θ - sin5θ - sin7θ).
2)
- 64 sin^7 x
= - 64 sin^4 x sin^3 x
= 4 (2 sin^2 x)^2 * (- 4 sin^3 x)
= 4 (1 – cos2x)^2 * (sin3x – 3sinx)
= 4 (1 – 2cos2x + cos^2 2x) * (sin3x – 3sinx)
= (2 – 4cos2x + 2cos^2 2x) * (2sin3x – 6sinx)
= (2 – 4cos2x + 1 + cos4x) * (2sin3x – 6sinx)
= (3 – 4cos2x + cos4x) * (2sin3x – 6sinx)
= 6sin3x – 18sinx – 8sin3x cos2x + 24sinx cos2x + 2sin3x cos4x - 6sinx cos4x
= 6sin3x – 18sinx – 4sin5x – 4sinx + 12sin3x – 12sinx + sin7x – sinx – 3sin5x + 3sin3x
= sin7x – 7sin5x + 21sin3x – 35sinx.

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