How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

Let $$ S = \sin{(a)} + \sin{(a+d)} + \cdots + \sin{(a+nd)}$$ Now multiply both sides by $\sin\frac{d}{2}$. Then you have $$S \times \sin\Bigl(\frac{d}{2}\Bigr) = \sin{(a)}\sin\Bigl(\frac{d}{2}\Bigr) + \sin{(a+d)}\cdot\sin\Bigl(\frac{d}{2}\Bigr) + \cdots + \sin{(a+nd)}\cdot\sin\Bigl(\frac{d}{2}\Bigr)$$

Now, note that $$\sin(a)\sin\Bigl(\frac{d}{2}\Bigr) = \frac{1}{2} \cdot \biggl[ \cos\Bigl(a-\frac{d}{2}\Bigr) - \cos\Bigl(a+\frac{d}{2}\Bigr)\biggr]$$ and $$\sin(a+d) \cdot \sin\Bigl(\frac{d}{2}\Bigr) = \frac{1}{2} \cdot \biggl[ \cos\Bigl(a + d -\frac{d}{2}\Bigr) - \cos\Bigl(a+d+\frac{d}{2}\Bigr) \biggr]$$

Then by doing the same thing you will have some terms cancelled out. You can easily see which terms are going to get Cancelled. Proceed and you should be able to get the formula.

I tried this by seeing this post. This has been worked for the case when $d=a$. Just take a look here:

• http://mathforum.org/library/drmath/view/72777.html

Writing $\cos x = \frac12 (e^{ix} + e^{-ix})$ will reduce the problem to computing two geometric sums.

Here's a trigonograph for $a = 0$ and $d = 2\theta$: