34.
ให้ $z=\cos\left(\dfrac{\pi}{9}\right)+i\sin\left(\dfrac{\pi}{9}\right)$
จะได้ว่า
$\sin 20^{\circ}=\dfrac{z-z^{-1}}{2i}$
$\sin 40^{\circ}=\dfrac{z^2-z^{-2}}{2i}$
$\sin 80^{\circ}=\dfrac{z^4-z^{-4}}{2i}$
และ $z^9=-1$
$(z^3+1)(z^6-z^3+1)=0$
$z^6=z^3-1$
$z^7=z^4-z$
$z^8=z^5-z^2$
$z^9=-1$
จึงได้
$\dfrac{1}{\sin^2 20^{\circ}}+\dfrac{1}{\sin^2 40^{\circ}}+\dfrac{1}{\sin^2 80^{\circ}}=\dfrac{-4}{(z-z^{-1})^2}+\dfrac{-4}{(z^2-z^{-2})^2}+\dfrac{-4}{(z^4-z^{-4})^2}$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-4\left[\dfrac{z^{14}+3z^{12}+3z^{10}+7z^8+3z^6+3z^4+z^2}{(z^8-1)^2}\right]$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-4\left[\dfrac{z^{14}+3z^{12}+3z^{10}+7z^8+3z^6+3z^4+z^2}{(-z^{-1}-1)^2}\right]$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-4\left[\dfrac{z^{16}+3z^{14}+3z^{12}+7z^{10}+3z^8+3z^6+z^4}{(z+1)^2}\right]$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-4\left[\dfrac{-(z^4-z)-3z^{5}-3z^{3}-7z+3(z^5-z^2)+3(z^3-1)+z^4}{(z+1)^2}\right]$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=-4\left[\dfrac{-3z^{2}-6z-3}{(z+1)^2}\right]$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=12$