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In the book I am reading I am asked to prove the following:

$$\sum_{n=1}^k{\cos n}<\frac{1}{2\sin{\frac{1}{2}}}-\frac{1}{2}$$

My attempt:

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

From here the series seems to telescope, but the thing is that I cannot find the patter to simplify it. After doing this I can get an expression that is equal to the initial sum, replace it in the orginal inequality and prove it, but right now I am stuck.

Any hint or ideas is welcome!

Samuel A. Morales
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    $\cos n=\dfrac{e^{in}+e^{-in}}{2}$, then sum geometrical series for $e^{in}$ and for $e^{-in}$ separately, then combine to $\sin$ or $\cos$ back, if needed. – Alexey Burdin Jul 02 '20 at 16:15
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    Is there anyway of doing it without complex analysis? The book has not talk about it yet so I guess we are not suppose to use it. Plus, the excersise is at the end of the telescoping series chapter. – Samuel A. Morales Jul 02 '20 at 16:16

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Hint: You made a small mistake: $$ 2\sin\frac12\cos n=\sin(n+\frac12)-\sin(n-\frac12). $$ After this correction the series will telescope.

user
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Hint...$$\sum_{n=1}^k{\cos(a+(n-1)b)=\frac{\cos [a+ (k-1)b]\sin\frac {kb}{2}}{\sin\frac b2}}$$ In This case $a=1$ and $b=1$,so, $$\sum_{n=1}^k{\cos(n)=\frac{\cos [\frac {k+1}{2}]\sin\frac {k}{2}}{\sin\frac 12}}$$