Summation Cosec Squared

The summation can be written as

$\displaystyle \begin{aligned} \frac{1}{\sin^2\left(\frac{\pi}{14}\right)}+\frac{1}{\sin^2\left(\frac{3\pi}{14}\right)}+\frac{1}{\sin^2\left(\frac{5\pi}{14}\right)} =2\left[\frac{1}{1-\cos\left(\frac{\pi}{7}\right)}+ \frac{1}{1-\cos\left(\frac{3\pi}{7}\right)}+\frac{1}{1-\cos\left(\frac{5\pi}{7}\right)}\right] \end{aligned}$

So now the question boils down to finding the value of $\dfrac{1}{1-\alpha}+\dfrac{1}{1-\beta}+\dfrac{1}{1-\gamma}$ , where $\alpha=\cos\left(\dfrac{\pi}{7}\right), \beta=\cos\left(\dfrac{3\pi}{7}\right),\gamma= \cos\left(\dfrac{5\pi}{7}\right)$ .

Now we find an equation whose roots are $\alpha,\beta,\gamma$ . For that we let $7\theta=\pi$ , so that $\cos 7\theta =-1$ .

Now we write the expansion of $\cos7\theta$ by using the identity $\cos7\theta+i\sin7\theta=(\cos \theta + i\sin \theta)^7$ and equating the real parts and then converting all the $\sin$ terms to $\cos$ terms by using $\sin^2\theta=1-\cos^2\theta$ .

Finally we get the expansion as $\cos7\theta=64\cos^7\theta-112\cos^5\theta+56\cos^3\theta-7\cos\theta=-1$ .

Notice that when we set $7\theta=\pi$ , the values of $\theta$ for which $\cos7\theta=-1$ will be $\dfrac{\pi}{7},\dfrac{3\pi}{7},\dfrac{5\pi}{7},\pi,\dfrac{9\pi}{7},\dfrac{11\pi}{7},\dfrac{13\pi}{7}$ , which are the $7$ roots of our equation. Now we remove the root $\theta=\pi$ by dividing by the factor $1+\cos\theta$ . (It must be factor since $\cos\theta=-1$ is a root).

So after dividing, we get our equation in $x$ as (where $x=\cos\theta$ ) , $64x^6-64x^5-48x^4+48x^3+8x^2-8x+1=0$ .

Also note that $\cos\left(\dfrac{\pi}{7}\right)=\cos\left(\dfrac{13\pi}{7}\right)$ are one and the same thing. It goes the same for other two pairs.

Hence there are three repeated roots. Hence the equation must be of the form $(x-\alpha)^2(x-\beta)^2(x-\gamma)^2$ . Therefore we take it's square root to get rid of repeated roots. So we get our new equation as $8x^3-4x^2-4x+1=0$ . Now for finding the sum $\dfrac{1}{1-\alpha}+\dfrac{1}{1-\beta}+\dfrac{1}{1-\gamma}$ , we can transform the roots by letting $t=\dfrac{1}{1-x}$ so that $x=\dfrac{t-1}{t}$ .

After substituting the value of $x$ , we get the final equation as $t^3-12t^2+20t-8=0$ . Therefore, the sum we seek is the sum of it's roots, which is $12$ . So we get our answer as $2\times 12=\boxed{24}$ .

Used TI-83 PLUS,and for accuracy calculated the three terms separately and then added.