Prove $2\cos5B+2\cos4B+2\cos3B+2\cos2B+2\cos B+1=\frac{\sin(11B/2)}{{\sin(B/2)}}$ Working through a book, I have stumbled upon a question I don't know how to solve:

Prove 
  $$2\cos(5B) + 2\cos(4B) + 2\cos(3B) + 2\cos(2B) + 2\cos(B) + 1 =  \dfrac{\sin\left(\frac{11B}2\right)}{{\sin\left(\frac B2\right)}}$$

(this is a smaller problem I have reduced the larger problem down from)
I don't know how to simply the cosine functions into a sine functions, as using the identity $\cos A + \cos B = 2\cos\left(\dfrac{A+B}2\right)\cos\left(\dfrac{A-B}2\right)$ simply gives me a result with even more cosine functions. I can't think of other identities that are helpful.
 A: Multiply both sides by $$\sin (B/2)$$
then use the formula  $$ 2\sin (a) \cos (b)= \sin(a+b) + \sin (a-b)$$
The result telescopes and you get the required identity. 
A: \begin{align}
& 1 + 2\cos B + 2\cos(2B) + 2\cos(3B) + 2\cos(4B) + 2\cos(5B) \\[8pt]
= {} & \cos(-5B) + \cos(-4B) + \cdots + \cos(0B) + \cdots + \cos(4B) + \cos(5B) \\
& \text{(since cosine is an even function)} \\[10pt]
= {} & \operatorname{Re} \big( e^{-5iB} + e^{-4iB} + \cdots + e^{0iB} + \cdots + e^{4iB} + e^{5iB} \big) \\[8pt]
= {} & e^{-5iB} + e^{-4iB} + \cdots + e^{0iB} + \cdots + e^{4iB} + e^{5iB} \\
& \text{(since the imaginary parts cancel out)} \\[10pt]
= {} & \text{a finite geometric series} \\[8pt]
= {} & \text{(first term)} \times \frac{1 - (\text{common ratio})^\text{number of terms}}{1 - (\text{common ratio})} \\[8pt]
= {} & e^{-5iB} \cdot \frac{1 - e^{11iB}}{1 - e^{iB}} = \frac{e^{-5iB} - e^{6iB}}{1 - e^{iB}} \\[10pt]
= {} & \frac{e^{-11iB/2} - e^{11iB/2}}{e^{-iB/2} - e^{iB/2}} \quad \left( \text{We multiplied by } \frac{e^{-iB/2}}{e^{-iB/2}}. \right) \\[10pt]
= {} & \frac{\sin(11B/2)}{\sin(B/2)}.
\end{align}
