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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.

Blue
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El_Mathematica
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    Multiply both sides by sin B/2 and then express cos nB sin B/2 as a difference of two sines. – almagest Dec 28 '19 at 21:23
  • In this case rather than "how to solve" one should say "how to prove". One solves problems; one solves equations; one evaluates expressions; one proves propositions; in particular one proves trigonometric identities. – Michael Hardy Dec 28 '19 at 22:04

2 Answers2

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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.

Mohammad Riazi-Kermani
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\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}

Michael Hardy
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