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So I'm working on proving Lagrange's trig identity

$$\sum_{k=0}^n\cos k\theta=\frac{1}{2}+\frac{\sin(2n+1)\frac{\theta}{2}}{2\sin\frac{\theta}{2}}$$

Skipping some steps I arrive here:

$$=\frac{[\cos(n+1)\theta-1][\cos\theta-1]+[\sin(n+1)\theta][\sin\theta]}{[\cos\theta-1]^2+\sin^2\theta}$$

$$=\frac{[\cos(n+1)\theta-1][{\cos\theta-1}]+[\sin(n+1)\theta][\sin\theta]}{-2(\cos\theta-1)}$$

$$\frac{[\cos(n+1)\theta-1]}{-2}+\frac{[\sin(n+1)\theta][\sin\theta]}{-2(\cos\theta-1)}$$

Both my book and other online sources all simply skip the trig manipulation required to finish proving the identity as if it's trivial. I honestly see no way to manipulate any of the above equations to the desired result and am wondering what I'm missing.

Leonidas Lanier
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  • Do you want to prove the result from the given expressions or a simple alternative proof would also suffice? – Math Lover Sep 06 '17 at 01:54
  • I'm aware there are proofs involving telescoping sums etc. but I'm asked to do it this way – Leonidas Lanier Sep 06 '17 at 01:54
  • I'm not sure about what you mean by "this" way. You may find the following expressions useful: $2\cos(A)\cos(B) = \cos(A+B) + \cos(A-B)$, $2\sin(A)\sin(B) = \cos(A-B) - \cos(A+B)$, and $2\sin(A)\cos(B) = \sin(A+B) + \sin(A-B)$. – Math Lover Sep 06 '17 at 02:06
  • By "this way" I meant $\Re\sum_{k=0}^ne^{ik\theta}$ – Leonidas Lanier Sep 06 '17 at 02:18

2 Answers2

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Note that $$\sum_{k=0}^{n}e^{i k \theta} = \frac{e^{i (n+1)\theta}-1}{e^{i \theta}-1} = \frac{e^{i(n+1)\theta/2}}{e^{i \theta/2}}\cdot\frac{e^{i (n+1)\theta/2}-e^{-i (n+1)\theta/2}}{e^{i \theta/2}-e^{-i\theta/2}} = e^{i n\theta/2}\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}.$$ Consequently, $$\Re\left(\sum_{k=0}^{n}e^{i k \theta}\right) = \sum_{k=0}^{n}\cos(k \theta) = \frac{\cos(n\theta/2)\sin((n+1)\theta/2)}{\sin(\theta/2)}.$$ The result immediately follows by noting that $$\sin(A)\cos(B) = \frac{\sin(A+B)+\sin(A-B)}{2}.$$

Math Lover
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Or try induction.

$\sum_{k=0}^n\cos k\theta=\frac{1}{2}+\frac{\sin(2n+1)\frac{\theta}{2}}{2\sin\frac{\theta}{2}} $ is true for $n=0$.

If this is true for $n$, then, putting $n+1$ for $n$, you want to prove $\sum_{k=0}^{n+1}\cos k\theta=\frac{1}{2}+\frac{\sin(2n+3)\frac{\theta}{2}}{2\sin\frac{\theta}{2}} $.

But, by the induction hypothesis, $\sum_{k=0}^{n+1}\cos k\theta =\sum_{k=0}^{n}\cos k\theta+\cos((n+1)\theta) =\frac{1}{2}+\frac{\sin(2n+1)\frac{\theta}{2}}{2\sin\frac{\theta}{2}}+\cos((n+1)\theta) $ so you want to prove $\frac{1}{2}+\frac{\sin(2n+1)\frac{\theta}{2}}{2\sin\frac{\theta}{2}}+\cos((n+1)\theta) =\frac{1}{2}+\frac{\sin(2n+3)\frac{\theta}{2}}{2\sin\frac{\theta}{2}} $ or $\sin\frac{(2n+1)\theta}{2}+2\sin\frac{\theta}{2}\cos((n+1)\theta) =\sin\frac{(2n+3)\theta}{2} $ or $2\sin\frac{\theta}{2}\cos((n+1)\theta) =\sin\frac{(2n+3)\theta}{2}-\sin\frac{(2n+1)\theta}{2} $.

Using the formula for $\sin(a)-\sin(b)$ will finish the proof.

marty cohen
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