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Do you know how to prove that $\displaystyle\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ using induction?

I have tried with $n = 1$ which gives $\cos \frac{x}{2} = \frac{\sin(nx)}{(2\sin1/2x)}$

I am not sure on how to expand with the trigonometric formulas.

With $n= p+1$ I get LHS: $\cos(2(n+1)-1)$ which I summaries to $\cos(2n+1)$ which should be $\cos 2n \cos 1-\sin 2n \sin1$ plus the RHS $\frac{\sin(nx)}{(2\sin 1/2x)}$

RHS $p+ 1 = \frac{\sin(n+1x)}{(2\sin 1/2x)}$

Any ideas on how to proceed would be very helpful.

zar
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addde
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2 Answers2

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If $$\sum_{r=1}^n\cos\dfrac{(2r-1)}2x=\dfrac{\sin nx}{2\sin\dfrac x2}$$ holds true for $n=m$

$$\sum_{r=1}^{m+1}\cos\dfrac{(2r-1)}2=\dfrac{\sin mx}{2\sin\dfrac x2}+\cos\dfrac{[2(m+1)-1]}2x$$

$$=\dfrac{\sin mx+2\sin\dfrac x2\cos\dfrac{(2m+1)}2x}{2\sin\dfrac x2}$$

Using Werner's formula, $2\sin\dfrac x2\cos\dfrac{(2m+1)}2x=\sin(m+1)x-\sin mx$

lab bhattacharjee
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  • Hi again

    I understand that you multiply wtih {2\sin\dfrac x2} to get to =\dfrac{\sin mx+2\sin\dfrac x2\cos\dfrac{(2m+1)}2x}{2\sin\dfrac x2}. But i dont understand the expansion using the Werners formula. Can you explain that to me or point to where i should start in order to learn ?

    Thank you for you answer lab.

     – addde Jul 03 '15 at 09:15
  • @addde, See http://mathworld.wolfram.com/WernerFormulas.html and http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html – lab bhattacharjee Jul 03 '15 at 11:23
  • are you using the first of the Wernerformulas to get to your answer? Seeing as you have 2 sin x/2 which could be 2sin a and 1 cos (2m+1) / 2 which could be cos B – addde Jul 05 '15 at 06:06
  • @addde, the third of the list in the link – lab bhattacharjee Jul 05 '15 at 14:42
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I see this is quite an old question but here is a solution that assumes only basic high school compound angle formulas:

If we assume that $\displaystyle\sum_{k=1}^n \cos\left(\frac{(2k-1)x}{2}\right)=\frac{\sin(nx)}{2\sin\left(\frac x 2\right)}$ then it follows that:

$$\begin{align} \sum_{k=1}^n \cos\left(\frac{(2k-1)x}{2}\right)+\cos\left(\frac{(2n+1}{2}\right)&=\frac{\sin(nx)}{2\sin\left(\frac x 2\right)}+\cos\left(\frac{(2n+1)x}{2}\right)\\ \sum_{k=1}^{n+1} \cos\left(\frac{(2k-1)x}{2}\right)&=\frac{\sin nx + 2\sin\left(\frac x 2\right)\cos\left(nx+\frac x 2\right)}{2\sin\left(\frac x 2\right)}\\ &=\frac{\sin nx + 2\sin\left(\frac x 2\right)\left[\cos nx\cos\left(\frac x 2\right)-\sin nx\sin\left(\frac x 2\right)\right]}{2\sin\left(\frac x 2\right)}\\ &=\frac{\sin nx \left[1-2\sin^2\left(\frac x 2\right)\right]+2\sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\cos nx}{2\sin\left(\frac x 2\right)}\\ &=\frac{\sin nx\cos x+\sin x\cos nx}{2\sin\left(\frac x 2\right)}\\ &=\frac{\sin(nx+x)}{2\sin\left(\frac x 2\right)}=\frac{\sin\left((n+1)x\right)}{2\sin\left(\frac x 2\right)} \end{align}$$

Richard Ambler
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