<s>[INST] Evaluate: $\csc^2\left(\frac{\pi}{9}\right)+\csc^2\left(\frac{2\pi}{9}\right)+\csc^2\left(\frac{4\pi}{9}\right)$ 
$$\csc^2\left(\frac{\pi}{9}\right)+\csc^2\left(\frac{2\pi}{9}\right)+\csc^2\left(\frac{4\pi}{9}\right) \;=\; \text{???}$$

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