Is \sin^3 x=\frac{3}{4}\sin x-\frac{1}{4}\sin 3x ?

cooopt392a0i

cooopt392a0i

Answered question

2022-02-25

Is sin3x=34sinx14sin3x ?

Answer & Explanation

Haiden Frazier

Haiden Frazier

Beginner2022-02-26Added 10 answers

de Moivre's formula says
cos(3x)+isin(3x)=(cos(x)+isin(x))3
=(cos3(x)3cos(x)sin2(x))+i(3cos2(x)sin(x)sin3(x))
=(4cos3(x)3cos(x))+i(3sin(x)4sin3(x))
Therefore,
cos(3x)=4cos3(x)3cos(x)
sin(3x)=3sin(x)4sin3(x) (1)
Solving (1) for sin3(x) yields
sin3(x)=34sin(x)14sin(3x)
flytandikqk

flytandikqk

Beginner2022-02-27Added 4 answers

You can use De Moivre's identity:
Let's Call:
z=cosx+isinx
1z=cosxisinx
Now subtracting both equations together, we get
2isinx=z1z
And we know that:
zn=(cisx)n=cis  nx
So:
2isinx=z1z
8isin3x=(z1z)3
Expanding the RHS:
8isin3x=z31z33(z1z)
8isin3x=2isin3x6isinx
sin3x=34sinx14sin3x

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