Q.357. Trigonometric identity

Question 357.
Prove that -64sin^7 θ = sin7θ - 7sin5θ + 21sin3θ - 35sinθ.

Answer 357.
Though I had not posted my answer on Yahoo, I have given it a place in my blog with my solution as the question is good and challenging.

- 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.

Link to YA!