Evaluating $\sum_{k=1}^{10}\left(\sin\frac{2k\pi}{11}+i\cos\frac{2k\pi}{11}\right)$

Question

Find the value of $$\sum_{k=1}^{10}\left(\sin\frac{2k\pi}{11}+i\cos\frac{2k\pi}{11}\right)$$

I have heard somewhere that this question can be done by $nth$ roots of unity or by vector algebra. But I'm looking for a solution involving just trigonometry. Is it possible to do it with trigonometry and applying almost no knowledge of vectors or $nth$ roots$?$

I have opened the summation and used the formulas of $$\sin x+\sin2x+\cdots+\sin nx$$ $$\cos x+\cos2x+\cdots+\cos nx$$ But I can't get any further. Also, if there are any other methods (other than I mentioned), I'll appreciate that.

Answer 1

A long comment rather than a solution. Since $\sin(-x)=-\sin(x)$ and $\cos(-x)=\cos(x)$, it holds $\sin(2\pi\frac{k}{11})=-\sin(2\pi\frac{11-k}{11})$ and $\cos(2\pi\frac{k}{11})=\cos(2\pi\frac{11-k}{11})$, whence the sum simplifies to $$2i\sum_{k=1}^5 \cos\left(2\pi\frac{k}{11}\right).$$

Answer 2

By properties of the polar representation of complex numbers, we can see that this is a sum of a geometric sequence. We can define:

$$a_k=\sin(\frac{2\pi k}{11})+i\cos(\frac{2\pi k}{11})=\cos(\frac{\pi}{2}-\frac{2\pi k}{11})+i\sin(\frac{\pi}{2}-\frac{2\pi k}{11})=e^{i(\frac{\pi}{2}-\frac{2\pi k}{11})}$$

And notice:

$$\frac{a_{k+1}}{a_k}=\frac{e^{i(\frac{\pi}{2}-\frac{2\pi (k+1)}{11})}}{e^{i(\frac{\pi}{2}-\frac{2\pi k}{11})}}=e^{i\frac{\pi}{2}-i\frac{\pi}{2}-i\frac{2\pi (k+1)}{11}+i\frac{2\pi k}{11}}=e^{-i \frac{2\pi}{11}}$$

This is a constant, so the sequence is indeed geometric with a commom ratio of $e^{-i \frac{2\pi}{11}}$.

To use the formula for the sum of the first $n$ terms of a geometric sequence we would also need to find:

$$a_1=e^{i(\frac{\pi}{2}-\frac{2\pi}{11})}=e^{i\frac{7\pi}{22}}$$

Finally, we can plug everything into the known sum formula:

$$\sum_{k=1}^{10}a_k=\fbox{$\frac{e^{i\frac{7\pi}{22}}(1-e^{-i \frac{10\cdot2\pi}{11}})}{1-e^{-i \frac{2\pi}{11}}}$}$$

And this can obviously be simplified to an expression looking like $a+bi$.

Answer 3

Let $x=\frac{\pi}{11}$ and so let

$$S=\cos2x+\cos4x+\cos6x+\cos8x+\cos10x\implies$$

$$2\cdot\sin x\cdot S=2\sin x\cos2x+2\sin x\cos4x+2\sin x\cos6x+2\sin x\cos8x+2\sin x\cos10x\implies$$

$$2\cdot\sin x\cdot S=\sin3x-\sin x+\sin5x-\sin3x+\sin7x-\sin5x+\sin9x-\sin7x+\sin11x-\sin9x$$

I applied $2\cos A\sin B=\sin(A+B)-\sin(A-B)$ So,

$$2\cdot\sin x\cdot S=\sin11x-\sin x$$

Undo the substitution, $$\sin11x=\sin\left(\frac{\pi}{11}\cdot 11\right)=\sin\pi=0$$

$$2\cdot\sin x\cdot S=0-\sin x \implies$$

$$S=\frac{-\sin x}{2\sin x} $$ or

$$S=\frac{-1}{2}$$

Answer 4

The crux of the matter is proving that

$\Sigma=[1]+[2]+[3]+[4]+[5]=-1/2,$

where for brevity $[n]$ is used for $\cos(2n\pi/11)$, or (with an eye towards later interpretation)

$[n]\overset{\text{def}}{=}\cos[1(2n\pi/11)].$

Take our expression for the sum

$\Sigma=[1]+[2]+[3]+[4]+[5]$

and multiply by $[1]$ to obtain

$[1]\Sigma=[1][1]+[1][2]+[1][3]+[1][4]+[1][5].$

Use the sum-product relation for cosines

$[m][n]=\frac12([m+n]+[m-n])$

together with the symmetry and periodicity relations

$[n]=[-n]=[11\pm n]$

to express $[1]\Sigma$ as a linear expression in the cosines:

$[1][1]=\frac12([2]+[0])=\frac12+\frac12[2]$

$[1][2]=\frac12([3]+[-1])=\frac12[1]+\frac12[3]$

et cetera, and thence

$[1]\Sigma=\frac12+\frac12[1]+[2]+[3]+[4]+[5].$

Similarly derive expressions for $[2]\Sigma, [3]\Sigma, [4]\Sigma, [5]\Sigma$. Note the nearly symmetrical nature of the results, which for completeness include the $[1]\Sigma$ product derived above:

$[1]\Sigma=\frac12+\frac12[1]+[2]+[3]+[4]+[5]$

$[2]\Sigma=\frac12+[1]+\frac12[2]+[3]+[4]+[5]$

$[3]\Sigma=\frac12+[1]+[2]+\frac12[3]+[4]+[5]$

$[4]\Sigma=\frac12+[1]+[2]+[3]+\frac12[4]+[5]$

$[5]\Sigma=\frac12+[1]+[2]+[3]+[4]+\frac12[5]$

And add everything up:

$([1]+[2]+[3]+[4]+[5])\Sigma=\frac52+\frac92([1]+[2]+[3]+[4]+[5]).$

Now the combinations $[1]+[2]+[3]+[4]+[5]$ can be replaced by $\Sigma$ and we have a quadratic equation for $\Sigma$:

$\Sigma^2-\frac92\Sigma-\frac52=0.$

Solving by standard methods gives two candidate roots $\Sigma=-\frac12$ and $\Sigma=5$. The latter is clearly impossible, or more precisely it's the sum we would get if we had chosen to render

$[n]\overset{\text{def}}{=}\cos[\color{blue}{0}(2n\pi/11)].$

instead of the intended

$[n]\overset{\text{def}}{=}\cos[\color{blue}{1}(2n\pi/11)].$

Thus the remaining root $-1/2$ must be the required sum value.

Answer 5

The easiest way is to observe that the expression is just a geometric series $$\sum_{k=1}^{10}z^k$$ where $z=e^{(2\pi /11)i}$. If you prefer to do the real and imaginary parts separately, you could use the forward difference operator $\Delta f(k)=f(k+1)-f(x)$ and $$\sum_{k=a}^b\Delta f(k)=f(b+1)-f(a)$$where$$\cos (k \theta)=\Delta(\sin ((k-1/2)\theta)/(2\cos(\theta /2)))$$ and $\theta=2\pi /11$ with a similar formula for the $\sin$ summation.