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I am a high school student and I am just learning about maths now, so please don't judge me if i ask stupid questions. anywho, I am trying to rewrite $$\sin (\pi x)$$ into something like: $$\sin(x) * [\text{something}]$$ I cant seem to find a general formula for that. Please help.

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    Unfortunately I don't think \there is a meaningful way you can write $\sin(\pi x)$ as a product involving $\sin(x)$. – 2017-01-21
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    we must know what is $x$ here? – 2017-01-21
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    a variable that is a member of the real numbers (wish i knew how to write that in a fancy mathematical notation using latex) – 2017-01-21

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I don't think you can express all of $\sin(ax)$ as a product involving $\sin(x)$ for all $a \in \mathbb{R}$ (Including $a=\pi$).

However, if $a \in \mathbb{Z}$ (integer), it can be written in the form you wanted.

Using the identities $\sin(2x)\equiv 2\sin(x)\cos(x)$, $\cos(2x)\equiv \cos^2(x)-\sin^2(x)$ and $\sin^2(x)+\cos^2(x)\equiv 1$, you can derive a majority of them. For example:

$$\sin(2x)=2\sin(x)\cos(x)$$ $$\sin(3x)=\sin(x)(2\cos(2x)+1)=3\sin(x)\cos^2(x)-\sin^3(x)=\sin(x)(-4\sin^2(x)+3)$$ $$\sin(4x)=2\sin(2x)\cos(2x)=4\sin(x)\cos(x)\cdot (\cos^2(x)-\sin^2(x))$$

This also applies for negative $a$, since $\sin(ax)$ is an odd function, i.e $f(-x)=-f(x)$: $$\sin(-2x)=-\sin(2x)=-2\sin(x)\cos(x)$$ $$\sin(-3x)=-\sin(3x)=-\sin(x)(2\cos(2x)+1)=-3\sin(x)\cos^2(x)-\sin^3(x)$$ $$\sin(-4x)=-\sin(4x)=-2\sin(2x)\cos(2x)=-4\sin(x)\cos(x)\cdot (\cos^2(x)-\sin^2(x))$$

Note: If you want, you can convert all $\cos(x)$ terms into $\sin(x)$ by substituting $\cos^2(x)\equiv 1-\sin^2(x)$.