Express $\sin(ax)$ in terms of $\sin(bx)$ or/and $\cos(bx)$ Is it possible to express $\sin(ax)$ in terms of $\sin(bx)$ or/and $\cos(bx)$ ?
That is, $\sin(ax)$ is a function of $\sin(bx)$ or/and $\cos(bx)$,
where x only exists inside $\sin(bx)$ or/and $\cos(bx)$.
$$\sin(ax) = f_1(\sin(bx)) \quad \vee \quad \sin(ax) = f_2(\cos(bx)) \quad \vee \quad \sin(ax) = f_3(\sin(bx), \cos(bx))$$
I want to find either of the functions $f_1$, $f_2$, $f_3$.
 A: Yes, it's possible with complex numbers. If you are allowed to use complex numbers, then use Euler's identity: $\Bbb e ^{\Bbb ix} = \cos x + \Bbb i \sin x$.
Then you get
$$
\sin ax = \frac {\Bbb e ^{\Bbb i ax} - \Bbb e ^{-\Bbb i ax}} {2 \Bbb i} = \frac {(\Bbb e^ {\Bbb i bx})^\frac a b - (\Bbb e ^{-\Bbb i bx})^\frac a b} {2 \Bbb i} .
$$
Using Euler's identity again for $\Bbb e ^{\Bbb i bx}$ and $\Bbb e  ^{-\Bbb i bx}$: $\Bbb e ^{\Bbb i bx} = \cos bx + \Bbb i \sin bx$ you get
$$\sin ax = \Bbb i \frac {(\cos bx - \Bbb i \sin bx)^\frac a b - (\cos bx + \Bbb i \sin bx)^\frac a b} 2 .$$
I also used the property of the imaginary unit $\frac 1 {\Bbb i} = -\Bbb i$.

To make my answer more useful, here is how we can evaluate $(\cos bx \pm \Bbb i \sin bx)^\frac a b$ for any real $a, b$
If $\cos bx \geq \sin bx$
$$(\cos bx \pm \Bbb i \sin bx)^\frac a b=\cos bx ^{ \frac a b } (1 \pm i \tan bx)^\frac a b$$
Which is binomial series and can be expanded.
If $\cos bx \leq \sin bx$
$$(\cos bx \pm \Bbb i \sin bx)^\frac a b=(-i \sin bx )^{ \frac a b } (1 \pm i \tan^{-1} bx)^\frac a b$$
A: If you restrict yourself to the real numbers, then it is not possible in general.
For example, take $a=1$ and $b=2$. Given any $x \in \mathbb{R}$ we have
$$\sin(x+\pi)=-\sin x$$
but since $\sin 2(x+\pi) = \sin 2x$ and $\cos 2(x+\pi) = \cos 2x$, no function of $\sin 2x$ and $\cos 2x$ can have this property.
