Trigonometric Inequality $\cos 1 +\cos2+\ldots +\cos n < 0.55$ can be solved with the help of Integrals?

Question

How can I prove for every $n \in \mathbb{N}$ $$\cos 1 +\cos2+\ldots +\cos n < 0.55$$

Any idea, any solution? Thanks!

EDIT

Can be solved this inequality with the help of integrals, because I met this exercise in the chapter entitled Integrals.

Thanks.

Answer 1

Since you asked for ideas, note that $e^{i\theta}=cos \theta + i \sin \theta$, so your sum is the real part of the geometric sum $$e^i+e^{2i}+\dots e^{ni}=\frac {e^{(n+1)i}-e^i}{e^i-1}$$

PS I've just followed David Mitra's comment - and this idea is more fully expanded in some of those answers. Worth working through once for yourself, though.

Answer 2

Call your sum $S(n)$. I'll use a trigonometric identity to turn $S(n)$ into a telescoping series and then find a good bound.

Using the identity $2\sin A \cos B = \sin(A+B)-\sin(A-B)$, we see that \begin{align} 2\sin(1/2)S(n) &= 2\sin(1/2)\cos1 + 2\sin(1/2)\cos2 + \cdots + 2\sin(1/2)\cos n \\ &= (\sin(3/2)-\sin(1/2))+(\sin(5/2)-\sin(3/2)) + \cdots + (\sin(n+1/2)-\sin(n-1/2))\\ &= \sin(n+1/2)-\sin(1/2). \end{align} Then $$S(n)=\frac{\sin(n+1/2)-\sin(1/2)}{2\sin(1/2)} =\frac{\sin(n+1/2)}{2\sin(1/2)}-\frac{1}{2}.$$ Since $\sin(n+1/2)<1$, we have $$S(n) < \frac{1}{2\sin(1/2)} - \frac{1}{2}=0.5429\ldots<0.55.$$