Okay, so heres what I mean by this... I know that:

$\displaystyle f(x) = sin(2\theta) = 2\cdot sin(\theta) \cdot cos(\theta)$


$\displaystyle g(x) = sin(3\theta) = 3 \cdot sin(\theta) - 4 \cdot sin^3(\theta)$

So, with that, I also know that:

$\displaystyle \frac{d}{d\theta} [ f(x) ] = 2 \cdot cos(2\theta) = 2 [cos^2(\theta) - sin^2(\theta)] $


$\displaystyle \frac{d}{d\theta} [ g(x) ] = 3 \cdot cos(3\theta) = 3 [4 \cdot cos^3(\theta) - 3 \cdot cos(\theta)] $

Now, is there some way to continue this process? To find some forumula for

$\displaystyle sin(n\theta) = h(\theta)$

simmilar to the forumulas above? And then use the derivitive of this sin function to find the forumula for the corresponding formula

$\displaystyle cos(n \theta) = j(\theta)$

? Like, for the examples above, $\displaystyle n = 2$ and $\displaystyle n =3$

Is there a general formula for this that we can use by differentiating it to get to the cosine version of said formula? Or is that not a valid method to use to get from the sine version to the consine version?