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sunmoonlight · Mar 26, 2024

How to do part d?

tywebb · Mar 26, 2024

$\text{Roots of }8x^3-6x-\sqrt3=0\text{ are }\alpha,\beta,\gamma$

$\begin{aligned}\textstyle\tan^2\frac{\pi}{18}+\tan^2\frac{11\pi}{18}+\tan^2\frac{13\pi}{18}&=\textstyle\sec^2\frac{\pi}{18}-1+\sec^2\frac{11\pi}{18}-1+\sec^2\frac{13\pi}{18}-1\\ &=\textstyle\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}-3\\ &=\textstyle\frac{(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)}{(\alpha\beta\gamma)^2}-3\\ &=\textstyle\frac{(-6/8)^2-2(\sqrt3/8)(0)}{(\sqrt3/8)^2}-3\\ &=9 \end{aligned}$

You can also check with calculator you get 9, not always possible as even the new surd-enabled calculators still have their limitations.

But checking on calculator in this case will work.

ivanradoszyce · Apr 2, 2024

Part (c) and (d) involves Vieta's Formula for cubics. This PDF may help is solving such problems. Created in Latex.

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