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 $i=\sqrt{-1}$.

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

$\cos^2(x)+\sin^2(x) = 1 \implies \cos(x) = \pm \sqrt{1-\sin^2(x)}$

this reduces to

$\sin(4x) = \pm 4 \sqrt{1-\sin^2(x)} \sin(x) (1 - 2 \sin^2(x))$