Summation of $\cos A+\cos(A+B)+ \ldots \cos(A+(n-1)B)$ How would I prove the following result? 
$$\cos A+\cos(A+B)+\ldots +\cos(A+ (n-1)B) = \frac{\sin\left(\frac{nB}{2}\right) \cos\left[A+\frac{(n-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$
 A: Hint: Use $\sum_{k = 1}^{n} \text{cos}(A+kB) = \text{Re}(\sum_{k=1}^{n}e^{i(A+kB)})$ and proceed with the sum of a geometric series.
A: Let $$ C=Cos(A)+Cos(A+B)+\cdots+Cos(A+(n-1)B) $$ and 
$$ S=Sin(A)+Sin(A+B)+\cdots+Sin(A+(n-1)B) $$ if $$i=\sqrt{-1}$$ Consider
$$C+iS=e^{iA}+e^{i(A+B)}+\cdots+e^{i(A+(n-1)B)}$$ $\implies$
$$C+iS=e^{iA}\left(1+e^{iB}+e^{i2B}+\cdots+e^{i(n-1)B}\right)$$ $\implies$
$$C+iS=e^{iA}\left(\frac{e^{inB}-1}{e^{iB}-1}\right)$$ $\implies$
$$C+iS=e^{i\left(A+(n-1)\frac{B}{2}\right)} \frac{Sin(\frac{nB}{2})}{Sin(\frac{B}{2})}$$ From this we get Real and Imaginary parts as the Required Results.
A: Induction on $n$.
Checking for $n=1$:
$$\cos A =\frac{\sin(\frac{B}{2})}{\sin(\frac{B}{2})}\cos A =\cos A  $$
Induction hypothesis: 
Suppose that the equality holds for $n=k$, thus
$$\cos A+\cos(A+B)+\ldots +\cos(A+ (k-1)B) = \frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$
Induction step:
Proving for $n=k+1$
$$\cos A+\cos(A+B)+\ldots +\cos(A+ (k-1)B)+\cos(A+kB) = \frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}+\cos(A+kB)$$
Now, we try to simplify the left right side:
$$\frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}+\cos(A+kB)$$
We have
$$\frac{\cos(A+kB)\sin (\frac{B}{2})+\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$
Using the identity:
$$\cos x \sin x = \frac{1}{2}[\sin(x+y)-\sin(x-y)]$$
$$\cos(A+kB)\sin (\frac{B}{2}) = \frac{1}{2}[\sin(A+kB+\frac{B}{2})-\sin(A+kB-\frac{B}{2})]$$
Or:
$$\cos(A+kB)\sin (\frac{B}{2}) = \frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})-\sin(A+\frac{B(2k-1)}{2})]$$
Plugging it:
$$\frac{\frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})-\sin(A+\frac{B(2k-1)}{2})]+\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}$$
Using the identity:
$$ \sin x\cos x = \frac{1}{2}[\sin(x+y)+\sin(x-y)]$$
We will get:
$$\frac{\frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})-\sin(A+\frac{B(2k-1)}{2})]+\frac{1}{2}[\sin(A+\frac{B(2k-1)}{2})+\sin(-A+\frac{B}{2})]}{\sin \left(\frac{B}{2}\right)}$$
After simplifying more:
$$\frac{\frac{1}{2}[\sin(A+\frac{B(2k+1)}{2})]+\frac{1}{2}[+\sin(-A+\frac{B}{2})]}{\sin \left(\frac{B}{2}\right)}$$
Using the identity:
$$\sin(x\pm y)=\sin x\cos y \pm \cos x \sin y $$
We will have:
$$\frac{\frac{1}{2}[\sin A \cos(\frac{B(2k+1)}{2})+\cos A\sin(\frac{B(2k+1)}{2})+\sin(\frac{B}{2})\cos(A)-\cos(\frac{B}{2})\sin(A)]}{\sin \left(\frac{B}{2}\right)}$$
Which equals to:
$$\frac{\frac{1}{2}[\sin( A)(\cos(\frac{B(2k+1)}{2})-\cos(\frac{B}{2}))+\cos(A)(\sin(\frac{B(2k+1)}{2})+\sin(\frac{B}{2}))}{\sin \left(\frac{B}{2}\right)}$$
Using the identities:
$$\cos x-\cos y=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})$$
$$\sin x+\sin y=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$$
We will get:
$$\frac{\frac{1}{2}[-2\sin( A)(\sin(\frac{B(k+1)}{2})\sin(\frac{Bk}{2}))+2\cos(A)(\sin(\frac{B(k+1)}{2})\cos(\frac{Bk}{2}))}{\sin \left(\frac{B}{2}\right)}$$
After simplifying:
$$\frac{-\sin( A)(\sin(\frac{B(k+1)}{2})\sin(\frac{Bk}{2}))+\cos(A)(\sin(\frac{B(k+1)}{2})\cos(\frac{Bk}{2}))}{\sin \left(\frac{B}{2}\right)}$$
and
$$\frac{\sin(\frac{B(k+1)}{2})[\cos(A)\cos(\frac{Bk}{2})-\sin(A)\sin(\frac{B(k+1)}{2})]}{\sin \left(\frac{B}{2}\right)}$$
Using the identity:
$$\cos x \cos y - \sin x \sin y=\cos(x+y)$$
We will get:
$$\frac{\sin(\frac{B(k+1)}{2})(\cos(A+Bk))}{\sin \left(\frac{B}{2}\right)}$$
Therefore
$$\cos A+\cos(A+B)+\ldots +\cos(A+ (k-1)B)+\cos(A+kB) = \frac{\sin\left(\frac{kB}{2}\right) \cos\left[A+\frac{(k-1)B}{2}\right]}{\sin \left(\frac{B}{2}\right)}+\cos(A+kB)$$
And we prove for $n=k+1$, thus the induction step completed, and the identity holds for any $n$.
