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Find the sum of following series:

$$1 + \frac{1}{4!} \cos 4\theta + \frac{1}{8!} \cos 8\theta + ...$$

My attempt:

I need hint to start. Thanks!

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  • $\begingroup$ both. i thought about bringing in $e^{i\theta}$ but that doesnt help much here $\endgroup$ – Foggy Aug 14 '15 at 16:38
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HINT $$\frac 12(\cosh(x)+\cos(x))=\frac 12\left(\sum_{k=0}^\infty \frac {x^{2k}}{(2k)!}+\sum_{k=0}^\infty \frac {(-1)^kx^{2k}}{(2k)!}\right)=\sum_{k=0}^\infty \frac {x^{4k}}{(4k)!}$$ and you have: $$\sum_{k=0}^\infty \frac{\cos(4k\theta)}{(4k)!}$$ and $\cos(x)=\Re(e^{ix})$

The answer turns out to be: $\dfrac 12 \left(\cos(\sin(\theta))\cosh(\cos(\theta))+ \cos(\cos(\theta))\cosh(\sin(\theta))\right)$

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  • $\begingroup$ answer is Real art of 1/2(cosh($e^{i\theta})+cos(e^{i\theta}$)) isnt it? $\endgroup$ – Foggy Aug 14 '15 at 16:52
  • $\begingroup$ @Foggy Yes it is. Use trig addition formulas, and the fact that $\cosh(ix)=\cos(x)$ and you'll get there. $\endgroup$ – GeorgSaliba Aug 14 '15 at 16:55
  • $\begingroup$ @GeorgSaliba How did you do that awesome thing with the answer? like, it appears when the cursor is hovering on it.. $\endgroup$ – Arkya Sep 3 '16 at 7:12

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