Evaluate: $\csc^2\left(\frac{\pi}{9}\right)+\csc^2\left(\frac{2\pi}{9}\right)+\csc^2\left(\frac{4\pi}{9}\right)$

Question

$$\csc^2\left(\frac{\pi}{9}\right)+\csc^2\left(\frac{2\pi}{9}\right)+\csc^2\left(\frac{4\pi}{9}\right) \;=\; \text{???}$$

$\bf{My\; Try::}$ Let $\displaystyle \frac{\pi}{9} = \theta\;,$ Then $9\theta = \pi\Rightarrow 6\theta = \pi-3\theta$

So $\sin (6\theta) = -\sin (3\theta)\Rightarrow 2\sin 3 \theta \cdot \cos 3 \theta+\sin 3 \theta =0$

So we get $\displaystyle \sin 3 \theta \cdot \left[2\cos 3 \theta +1\right] =0$

So we get $\sin 3\theta = 0$ or $2\cos 3 \theta+1=0$

Now i did not understand how can I convert into $\sin^2$ form, Help me

Thanks

Answer 1

Like my answer here in Sum of tangent functions where arguments are in specific arithmetic series

$$\tan9x=\dfrac{\binom91t-\binom93t^3+\binom95t^5-\binom97t^7+t^9}{\cdots}$$ where $t=\tan x$

If $\tan9x=0,9x=n\pi$ where $n$ is any integer

So, the roots of $$\binom91t-\binom93t^3+\binom95t^5-\binom97t^7+t^9=0$$ are $\tan\dfrac{r\pi}9$ where $n\equiv0,\pm1,\pm2,\pm3,\pm4\pmod9$

As $\tan0=0,$ the roots of $$\binom91-\binom93t^2+\binom95t^4-\binom97t^6+t^8=0$$ are $\tan\dfrac{r\pi}9$ where $r\equiv\pm1,\pm2,\pm3,\pm4\pmod9$

Putting $c=1/t,$ the roots of $$\binom91c^8-\binom93c^6+\binom95c^4-\binom97c^2+1=0$$ are $\cot\dfrac{r\pi}9$ where $r\equiv\pm1,\pm2,\pm3,\pm4\pmod9$

Putting $c^2=d,$ the roots of $$\binom91d^4-\binom93d^3+\binom95d^2-\binom97d+1=0$$ are $\cot^2\dfrac{r\pi}9$ where $r\equiv1,2,3,4\pmod9$

$$\sum_{r=1}^4\cot^2\dfrac{r\pi}9=\dfrac{\binom93}{\binom91}=?$$

But $r=3\implies\cot^2\dfrac{3\pi}9=3$

Use $\csc^2u=1+\cot^2u$

Can you take it from here?

Answer 2

As a partial answer, finding the trig value for $\sin\left(\frac{\pi}{9}\right)$ is discussed here: http://mathworld.wolfram.com/TrigonometryAnglesPi9.html