How to prove \(\displaystyle{\sin{{3}}}\theta={3}{\sin{\theta}}-{4}{{\sin}^{{3}}\theta}\)

Adison Palmer

Adison Palmer

Answered question

2022-04-06

How to prove sin3θ=3sinθ4sin3θ

Answer & Explanation

Raina Blackburn

Raina Blackburn

Beginner2022-04-07Added 10 answers

We will use the following trigonometric formulas:
sin(x+y)=sinxcosy+sinycosx
sin(2x)=2sinxcosx
cos(2x)=12sin2x
cos2x=1sin2x
So
sin(2θ+θ)=sin(2θ)cosθ+sinθcos(2θ)
=2sinθcos2θ+sinθ(12sin2θ)
=2sinθ(1sin2θ)+sinθ2sin3θ
=3sinθ4sin3θ
lildeutsch11xq2j

lildeutsch11xq2j

Beginner2022-04-08Added 13 answers

sin3θ=sin(θ+2θ)
=sinθcos2θ+cosθsin2θ
=sinθ(cos2θsin2θ)+2cos2θsinθ
=3cos2θsinθsin3θ
3(1sin2θ)sinθsin3θ
=3sinθ4sin3θ

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