PROVE THAT:- sin 5 theta=5 sin theta - 20 sin 3 theta + 16 sin 5 theta
Answers
Answer:
hence proved.
Step-by-step explanation:
et 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
Step-by-step explanation:
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Prove:
sin5θ=16sin
5
θ−20sin
3
θ+5sinθ
Hard
Solution
verified
Verified by Toppr
sin5A=sin(3A+2A)
=sin3Acos2A+cos3Asin2A
=(3sinA−4sin³A)(1–2sin²A)+cos(2A+A)sin2A
=3sinA−10sin³A+8sin
5
A+[cos2AcosA−sin2AsinA]sin2A
=3sinA−10sin³A+8sin
5
A+[(1–2sin²A)cosA−2sin²AcosA]2sinAcosA
=3sinA−10sin³A+8sin
5
A+[cosA−4sin²AcosA]2sinAcosA
=3sinA−10sin³A+8sin
5
A+2sinAcos²A−8sin³Acos²A
=3sinA−10sin³A+8sin
5
A+2sinA(1−sin²A)−8sin³A(1−sin²A)
=3sinA−10sin³A+8sin
5
A+2sinA−2sin³A−8sin³A+8sin
5
A
=5sinA−20sin³A+16sin
5
A[Proved]