<s>[INST] How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:
$$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$
There is a slight difference in case of $\sin$, which is:
$$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2  a + (n-1)\cdot d}{2}\biggr)$$ 
How do we prove the above two identities?
 [/INST] This is similar to the currently accepted answer, but more straightforward. You can use the trig identity
\begin{equation*}
    \sin(\alpha + \beta) - \sin(\alpha - \beta) = 2\sin \beta \cos \alpha.
\end{equation*}
Let $a_n = a + 2dk$ be an arithmetic sequence of difference $2d$, and set $b_n
= a_n - d = a + d(2k - 1)$. Note that $\{b_n\}$ is also an arithmetic sequence
of difference $2d$, hence $a_n + d = b_n + 2d = b_{n + 1}$. Therefore
\begin{equation*}
    2 \sin d \cos a_n = \sin(a_n + d) - \sin(a_n - d) = \sin b_{n + 1} - \sin b_n.
\end{equation*}
Summing both sides from $0$ to $n$ yields
\begin{align*}
    2 \sin d \sum_{k = 0}^n \cos a_k &= \sin b_{n + 1} - \sin b_0 \\
                                     &= \sin(a + d(2n + 1)) - \sin(a - d).
\end{align*}
From our original trig identity,
\begin{equation*}
    2\sin((n + 1)d) \cos(a + nd) = \sin(a + d(2n + 1)) - \sin(a - d).
\end{equation*}
Thus, if $\sin d \neq 0$, we can rewrite our result as
\begin{equation*}
    \sum_{k = 0}^n \cos (a + 2dk) = \frac{\sin((n + 1)d) \cos(a + nd)}{\sin d}.
\end{equation*}
This is OP's formula with $2d$ and $n$ instead of $d$ and $n - 1$. A similar process will yield the formula for $\sum_{k = 0}^n \sin(a + 2dk)$.
</s>