Finding value of $\displaystyle \bigg(\cot \frac{\pi}{18}-3\cot \frac{\pi}{6}\bigg)\cdot \bigg(\csc \frac{\pi}{9}+2\cot \frac{\pi}{9}\bigg)$
Try: $$\cot \frac{\pi}{18}\csc \frac{\pi}{9}-3\sqrt{3}\csc \frac{\pi}{9}+2\cot \frac{\pi}{18}\cot\frac{\pi}{9}-6\sqrt{3}\cot \frac{\pi}{9}$$
could some help me how can i simplify it,thanks
$$\cot x-3\cot3x=\dfrac{\cos x\sin3x-3(\cos3x\sin x)}{\sin x\sin3x}$$
Again,
\begin{align} 2\cos x\sin3x-3(2\cos3x\sin x) &=\sin4x+\sin2x-3(\sin4x-\sin2x)\\[4px] &=4\sin2x-2\sin4x \\[4px] &=4\sin2x(1-\cos2x)\\[4px] &=4\sin2x(2\sin^2x) \end{align}
Finally, \begin{align} \csc2x+2\cot2x &=\dfrac{1+2\cos2x}{\sin2x}\\[4px] &=\dfrac{1+2(1-2\sin^2x)}{\sin2x}\\[4px] &=\dfrac{\sin3x}{\sin x\sin2x} \end{align} for $\sin x\ne0$ using $\sin3x$ formula.
Can you identify $x$ here?
