Evaluating Trigonometric Expression

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SUMMARY

The evaluation of the expression $$\tan^2 20^{\circ}+\tan^2 40^{\circ}+\tan^2 80^{\circ}$$ can be derived from the roots of the cubic equation $$t^3−3√3t^2−3t+√3=0$$, where the roots correspond to $$\tan 20^{\circ}, -\tan 40^{\circ},$$ and $$\tan 80^{\circ}$$. By substituting $$t = x^{1/2}$$, the equation transforms into $$x^{3/2} - 3√3x - 3x^{1/2} + √3=0$$. The final cubic equation $$x^3 - 33x^2 + 27x - 3 = 0$$ reveals that the sum of the roots, which is $$\tan^2 20^{\circ} + \tan^2 40^{\circ} + \tan^2 80^{\circ}$$, equals 33.

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  • Familiarity with polynomial equations and their roots
  • Knowledge of substitution methods in algebra
  • Ability to manipulate and solve cubic equations
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anemone
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Without the use of a calculator, evaluate $$\tan^2 20^{\circ}+\tan^2 40^{\circ}+\tan^2 80^{\circ}$$.
 
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I won't cheat by repeating the solutions that I found here. But I previously found that there is also a neat answer for $\tan^210^\circ + \tan^250^\circ + \tan^270^\circ$, which can be found by the same methods.
 
anemone said:
Without the use of a calculator, evaluate $$\tan^2 20^{\circ}+\tan^2 40^{\circ}+\tan^2 80^{\circ}$$.

In the link provided by opalgs above

the roots of the equation t^3−3√3t^2−3t+√3=0

will be tan20,tan(−40)=−tan40,tan80

that is of f(t) =t^3−3√3t^2−3t+√3=0

and we shall construct an equation whose roots are
tan^2 20,tan^2 40 tan^2 80

shall be f(x^(1/2) = 0)

putting t = x^(1/2) we get

so x^(3/2) - 3√3x−3x^(1/2) +√3=0

or x^(3/2) - 3 x^(1/2) = 3√3x -√3
or
√x(x-3) = √3(3x -1)

square both sides to get

x(x-3)^3 = 3(3x-1)^2
or x(x^2-6x+ 9) = 3(9x^2 - 6x + 1)

or x^3 - 33 x^2 + 27x - 3 = 0

as it is cubic roots are tan^2 20,tan^2 40 tan^2 80
and sum of roots = - coefficient of x^2 or 33
 

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