<s>[INST] How to prove that $\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$ How to prove this trigonometric identities ?
$$\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$$
Thank you.
 [/INST] We have to check:
$$ -\tan\frac{\pi}{9}+\cot\frac{2\pi}{9}=4\cos\frac{4\pi}{9}\cot\frac{2\pi}{9} $$
hence, by multiplying both sides by $\sin\frac{2\pi}{9}=2\sin\frac{\pi}{9}\cos\frac{\pi}{9}$:
$$ -2\sin^2\frac{\pi}{9}+\cos\frac{2\pi}{9} = 4\cos\frac{4\pi}{9}\cos\frac{2\pi}{9} $$
so we just have to check that $x=\cos\frac{2\pi}{9}$ is a solution of:
$$ 2x-1 = 4x(2x^2-1) $$
or a root of:
$$ p(x) = 8x^3-6x+1 = 2\cdot T_3(x)-1, \tag{1}$$
where $T_3(x)=4x^3-3x$ is a Chebyshev polynomial. To check that $\cos\frac{2\pi}{9}$ is a root of the RHS of $(1)$ is trivial, since $T_3(\cos\alpha)=\cos(3\alpha)$:
$$ 2\cdot T_3\left(\cos\frac{2\pi}{9}\right)-1 = 2\cdot\cos\frac{2\pi}{3}-1 = 0, $$
hence we're done.
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