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

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


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

So, with that, I also know that:

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


\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

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

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

? Like, for the examples above, n = 2 and 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?