Question

Sin(4x) in the term of just x
Please help!!

Answer 1

I think you mean in terms of
`\sin(x)`?

Recall Euler's identity

`e^(ix) = \cos(x) + i \sin(x)`

and de Moivre's theorem

`\left(e^(ix)\right)^n = \left(\cos(x) + i \sin(x)\right)^n = \cos(nx) + i \sin(nx) = e^(inx)`

where
`i=√(-1)`.

It follows that

`\sin(4x) = \mathrm{Im}\left(\cos(x) + i \sin(x)\right)^4`

By the binomial theorem, expanding the right side gives

`\cos^4(x) + 4i \cos^3(x) \sin(x) - 6\cos^2(x) \sin^2(x) - 4i \cos(x) \sin^3(x) + \sin^4(x)`

and so

`\sin(4x) = 4\cos^3(x) \sin(x) - 4 \cos(x) \sin^3(x)`

We can factorize this as

`\sin(4x) = 4 \cos(x) \sin(x) \left(\cos^2(x) - \sin^2(x)\right)`

and using the Pythagorean identity

`\cos^2(x)+\sin^2(x) = 1 \implies \cos(x) = \pm √(1-\sin^2(x))`

this reduces to

`\sin(4x) = \pm 4 √(1-\sin^2(x)) \sin(x) (1 - 2 \sin^2(x))`