Question

Prove that 16sin^5θ − 20sin^3θ + 5sinθ = sin5θ

Answer 1

Step-by-step explanation:

Let theta = x

LHS sin5x

=sin(3x+2x)

=sin3x.cos2x+cos3x.sin2x

=(3sinx-4sin^3x).(1–2sin^2x)+(4cos^3x-3cosx).(2sinx.cosx)

=(3sinx-4sin^3x)(1–2sin^2x)+(4cos^4x-3cos^2x)(2sinx).

=(3sinx-4sin^3x)(1–2sin^2x)+cos^2x.(4cos^2x-3).(2sinx)

=3sinx-4sin^3x-6sin^3x+8sin^5x+(1-sin^2x).(4–4sin^2x-3).(2sinx)

=8sin^5x-10sin^3x+3sinx+(2sinx-2sin^3x)(1–4sin^2x).

=8sin^5x-10sin^3x+3sinx+2sinx-2sin^3x-8sin^3x+8sin^5x

=16sin^5x-20sin^3x+5sinx proved

Answer 2

See the picture above..

Hope it helps you..