$\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+.....+\cos(\alpha+(n-1)\beta)=0 $

Question

If each side of a regular polygon of $n$ sides subtend an angle $\alpha$ at the center of the polygon and each exterior angle of the polygon is $\beta$,then prove that $\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+.....+\cos(\alpha+(n-1)\beta)=0 $

Since this is a regular polygon.Therefore,each $\alpha=\frac{2\pi}{n}$ and since each external angle is $\beta$.So by geometry,$\alpha=\beta.$

$\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+.....+\cos(\alpha+(n-1)\beta)=\frac{\cos\frac{n\beta}{2}}{\cos\frac{\beta}{2}}\cos\frac{2\alpha+(n-1)\beta}{2}$

Now putting $\alpha=\beta=\frac{2\pi}{n}$ does not give me answer.What mistake did i make?

Answer 1

The mistake is in the summation formula. Here is a correct version:

$$\sum_{j=0}^{n-1}\cos (\alpha+\beta j)=\frac{ \sin \left(\frac{n \beta }{2}\right) }{\sin \left(\frac{\beta }{2}\right)}\cos \left(\alpha+\frac{1}{2} (n -1 )\beta\right).$$

This clearly helps to establish your result.

Answer 2

Obviously, the displayed "identity" is wrong. If you put $\alpha=\beta=0$, the left hand side is $n$, but the right hand side is $1$.

Edit. As pointed out by Math-fun in the other answer, you have apparently memorised the sum-of-angle formula wrongly. Alternatively, by De Moivre's formula, the sum in question is equal to the real part of $$ S=e^{i\alpha}(1+e^{i\beta}+e^{2i\beta}+\ldots+e^{(n-1)i\beta}). $$ Sum up the geometric sequence and put $\beta=\frac{2\pi}n$, you will see that $S$ and in turn its real part are zero.

Answer 3

Notice, in a regular polygon with $n$ number of side,

the angle subtended by each side at the center of the polygon $$\alpha=\frac{2\pi}{n}$$

each exterior angle of the polygon $$\beta=\pi-\frac{(n-2)\pi}{n}=\frac{2\pi}{n}$$

$$\implies \alpha=\beta=\frac{2\pi}{n}$$ Now, we have $$\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ldots +\cos(\alpha+(n-1)\beta)$$ setting $\beta=\alpha$ $$=\cos \alpha+\cos(\alpha+\alpha)+\cos(\alpha+2\alpha)+\ldots +\cos(\alpha+(n-1)\alpha)$$ $$=\cos \alpha+\cos(2\alpha)+\cos(3\alpha)+\ldots +\cos(n\alpha)$$ $$=\frac{1}{2}\left[\sin n\alpha\cot\frac{\alpha}{2}+\cos n\alpha-1\right]$$ Setting $\alpha=\frac{2\pi}{n}$ $$=\frac{1}{2}\left[\sin n\left(\frac{2\pi}{n}\right)\cot\frac{1}{2}\left(\frac{2\pi}{n}\right)+\cos n\left(\frac{2\pi}{n}\right)-1\right]$$ $$=\frac{1}{2}\left[\sin 2\pi\cot\frac{\pi}{n}+\cos 2\pi-1\right]$$ $$=\frac{1}{2}\left[0+1-1\right]=0$$ Hence, proved that

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ldots +\cos(\alpha+(n-1)\beta)=\color{blue}{0}}}$$