Is there a simple method to algebraically take a value outside of a trigonometric function argument Lets say I have a function $\sin(\frac{11\omega}{2})$ and I want to make the argument $\sin(\frac{\omega}{2})$,
I know that I cannot simply do this: $11*\sin(\frac{\omega}{2})$
So how would I go about removing the a constant from inside of a trig function?
 A: Using DeMoivre's Theorem:
$$\left(\cos \theta +i\sin \theta \right)^n=\cos n\theta +i\sin n\theta $$
We observe that the imaginary parts must be equal.
Case 1:n is odd.
$$\sin \left(n\theta \right)=\sum _{k=0}^{\frac{\left(n-1\right)}{2}}\binom{n}{2k+1}\sin ^{2k+1}\left(\theta \right)\cos ^{\left(n-\left(2k+1\right)\right)}\left(\theta \right)$$
Case 2: n is even
$$\sin \left(n\theta \right)=\sum _{k=1}^{\frac{\left(n\right)}{2}}\binom{n}{2k-1}\sin ^{2k-1}\left(\theta \right)\cos ^{\left(n-\left(2k-1\right)\right)}\left(\theta \right)$$
A: You can trade $\cos nx$ and $\sin nx$ for polynomials in $\cos x,\sin x$. Similarly, you can trade the powers $\cos^n x,\sin^n x$ for linear combinations of $\cos kx,\sin kx$.
I am not sure that this will help you.
A: Building on @YvesDaoust 's answer, and from a section of Wikipedia's trigonometric identities page, we have:
$$\sin{(a \pm b)} = \sin{(a)}\cos{(b)} \pm \cos{(a)}\sin{(b)} \tag{1}$$
$$\cos{(a \pm b)} = \cos{(a)}\cos{(b)} \mp \sin{(a)}\sin{(b)} \tag{2}$$
so that you could split the trig that way.  These, especially (1), can be used to derive the double angle formula
$$\sin{(2a)} = 2\sin{(a)}\cos{(a)} \tag{3}$$
You can also consider the multiple angle formula found on the same page here.  It's basically an extension of (3).
