How can we sum up $\sin^m$ and $\cos^m$ series when the angles are in arithmetic progression?
- Does an identity exist similar to 1.1 and 1.2 for $m>1$?, and
- Does an approximate or estimate exist?
For example here is the sum of $\cos$ series when $m=1$:
$$\sum_{k=0}^{n-1}\cos^m (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr) \,\,\,(1.1)$$
There is a slight difference in case of $\sin$, which is: $$\sum_{k=0}^{n-1}\sin^m (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2 a + (n-1)\cdot d}{2}\biggr) \,\,\,(1.2)$$
As you can see, in the case of $m=1$ an exact identity exists that is simple enough.
Once you have a formula for $\cos^m$ you get a formula for $\sin^m$ using $\sin(x)=\cos(\pi/2-x).$ An arithmetic sequence subtracted from $\pi/2$ is still an arithmetic series.
So we only need to think about $\cos^m.$
For $m=2,$ you can use that $$\cos^2(x)=\frac12(\cos(2x)+1).\tag{2}$$
For $m=3,$ you can use: $$\cos^3(x)=\frac14(\cos(3x)+3\cos(x))\tag{3}.$$
We can use these formulae to get formula for $m=2,3.$
For example, for $m=2$ you get:
$$\sum\cos^2=\frac{n}2+\frac12\frac{\sin nd}{\sin d}\cos(2a+(n-1)d)$$
The formula for $\sin^2$ is easy to derive from this:
$$\sum\sin^2=\frac{n}2-\frac12\frac{\sin nd}{\sin d}\cos(2a+(n-1)d)$$
For any $m,$ there is such a formula like (2) and (3), namely for $m$ odd:
$$\cos^m(x)=\frac1{2^{m-1}}\left(\sum_{k=0}^{\lfloor m/2\ \rfloor}\binom{m}{k}\cos(m-2k)x\right)$$
When $m$ is even, you have to adjust the the constant term:
$$\cos^m(x)=\frac1{2^{m-1}}\left(\frac12\binom{m}{m/2}+\sum_{k=0}^{ m/2-1}\binom{m}{k}\cos(m-2k)x\right)$$