Finding a closed form for $\cos{x}+\cos{3x}+\cos{5x}+\cdots+\cos{(2n-1)x}$

Question

We have to find

$$g(x)=\cos{x}+\cos{3x}+\cos{5x}+\cdots+\cos{(2n-1)x}$$

I could not get any good idea .

Intialy I thought of using

$$\cos a+\cos b=2\cos(a+b)/2\cos (a-b)/2$$

Answer 1

Following @TheDeadLegend's answer I found this telescoping technique. Turns out that you need a similar identity to make it work:

$$ \sin(\alpha + \beta) - \sin(\alpha - \beta) =2\cos \alpha \sin \beta$$

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\begin{align} g(x) &= \sum_{k=1}^n \cos(2k-1)x \\ &= \frac{1}{2\sin x}\sum_{k=1}^n 2\cos(2k-1)x \cdot \sin x \\ &= \frac{1}{2\sin x}\sum_{k=1}^n \left[\sin 2kx - \sin2(k-1)x \right] \\ &= \frac{1}{2\sin x}(\sin 2nx-0) = \frac{\sin 2nx}{2\sin x}. \end{align}

Answer 2

Let $z=\cos\theta+i\sin\theta$ i.e. $z=e^{i\theta}$

Your sum:$$e^{i\theta}+e^{3i\theta}+e^{5i\theta}+...e^{(2n-1)i\theta}$$

This is a GP with common ratio $e^{2i\theta}$

Therefore sum is $$\frac{a(r^n-1)}{r-1}$$ $$\frac{e^{i\theta}(e^{2ni\theta}-1)}{e^{2i\theta}-1}$$ $$\frac{(\cos \theta+i\sin\theta)(\cos(2n\theta)+i\sin\theta-1)}{\cos(2\theta)+i\sin(2\theta)-1}$$

Computing it's real part should give you the answer

Acknowledgement:Due credits to @LordShark Idea