Math, asked by IntelligentAshish, 4 months ago

Prove that
16 \sin^{5}{\theta} - 20\sin^{3}{\theta}  + 5\sin{\theta}  = \sin5{\theta}


ranitkarmaka: hi
IntelligentAshish: if you can't then why are you wasting my time
IntelligentAshish: 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

Answers

Answered by ranitkarmaka
1

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IntelligentAshish: A mad fellow
IntelligentAshish: 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
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