<s>[INST] A "simple" trigonometric computation... Show that $$\left(1+4\sin^2\frac{\pi}{18}\right) \left(1+4\sin^2\frac{\pi}{6}\right) \left(1+4\sin^2\frac{5\pi}{18}\right)\left(1+4\sin^2\frac{7\pi}{18}\right)=34.$$
Any ideas about how to tackle this?
 [/INST] Got it!

Consider $T(x)=256X^4-576X^3+432X^2-120X+9$.
Then $\cos^2\frac{\pi}{18}$, $\cos^2\frac{\pi}{6}$, $\cos^2\frac{5\pi}{18}$, $\cos^2\frac{7\pi}{18}$ are its roots.
$34$ is then found using Viete's formulas.
$T$ is derived from the 9-th Chebyshev polynomial.

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