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From playing around with a graph, I noticed that:

$$ \lim_{a \to \infty} \sum_{n=0}^{a} \left( \sin \left( x-\frac{\pi }{2} \right) \right)^n \cos \left( x-\frac{\pi }{2} \right) \approx \tan \left( 0.5x \right) $$

Can anyone prove this?

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    $\begingroup$ Do you know how to sum a geometric progression? $\endgroup$ – Mark Bennet Aug 7 '15 at 15:47
  • $\begingroup$ @MarkBennet, it is written with the argument of sine raised to a power. So it's not geometric. $\endgroup$ – Paul Aug 7 '15 at 15:52
  • $\begingroup$ @Paul Actually, the notation is ambiguous... It'd make sense for the OP to clarify this point. $\endgroup$ – Clement C. Aug 7 '15 at 15:54
  • $\begingroup$ @Paul Then the title of the post needs to be changed. $\endgroup$ – Mark Bennet Aug 7 '15 at 15:54
  • $\begingroup$ @Paul that is actually a good question - on the other hand I am pretty sure that not the argument was meant but rather the whole function (otherwise there would be no convergence whatsoever be possible) $\endgroup$ – user190080 Aug 7 '15 at 15:55
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$$\sin \left( x-\frac\pi2\right)=-\sin \left(\frac\pi2-x\right)=-\cos x$$

$$\cos \left( x-\frac\pi2 \right)=\cos \left(\frac\pi2-x\right)=\sin x$$

$$S=\sum_{n=0}^a\sin \left( x-\frac\pi2\right)^n\cos \left(x-\frac\pi2\right) $$

$$=\sin x\sum_{n=0}^a(-\cos x)^n$$ $$=\sin x\cdot\dfrac{1-(-\cos x)^{a+1}}{1+\cos x}$$

Now for real $x,|\cos x|\le1$ and for $x\ne2m\pi$ where $m$ is any integer, $|-\cos x|<1$

$$\implies\lim_{a\to\infty}(-\cos x)^{a+1}\to0$$

$$\implies\lim_{a\to\infty}S=\dfrac{\sin x}{1+\cos x}=\dfrac{2\sin\dfrac x2\cos\dfrac x2}{2\cos^2\dfrac x2}=?$$

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