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!
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, $$\sum_{n=1}^k{\cos(n)=\frac{\cos [\frac {k+1}{2}]\sin\frac {k}{2}}{\sin\frac 12}}$$