MHB Find the sum of three trigonometric terms

[anemone]

Evaluate $$\tan^4 10^\circ+\tan^4 50^\circ+\tan^4 70^\circ$$ without the help of a calculator.

 

[juantheron]

Spoiler

Let $\tan^2 10^\circ = x_{1}$ and $\tan^2 50^\circ =x_{2}$ and $\tan^2 70^\circ = x_{3}$

Then $\displaystyle \tan^2(3\cdot 10^\circ ) = \frac{1}{3}$ and $\displaystyle \tan^2(3\cdot 50^\circ ) = \frac{1}{3}$ and $\displaystyle \tan^2(3\cdot 70^\circ ) = \frac{1}{3}$

Using $\displaystyle \tan (3x) = \frac{3\tan x-\tan^3 x}{1-3\tan^2 x} = \frac{1}{\sqrt{3}}$

So $\displaystyle \frac{1}{3} = \left(\frac{3\tan x-\tan^3 x}{1-3\tan^2 x}\right)^2\Rightarrow (1-3\tan^2 x)^2 = 3(3\tan x-\tan^3 x)^2$

So $\displaystyle 1+9\tan^4 x-6\tan^2 x = 27\tan^2 x+3\tan^6 x-18\tan^4 x$

So $\displaystyle 3\tan^6 x-27\tan^4 x+33\tan^2 x-1=0$

has a roots $x_{i}=\tan^2(\theta)\;,$ Where $\theta = 10^\circ\;,50^\circ\;,70^\circ$

So we can say $x_{1}\;,x_{2}\;,x_{3}$ are the roots of $3x^3-27x^2+33x-1=0$

So $\displaystyle x_{1}+x_{2}+x_{3} = 9$ and $x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1} = 11$

So Using Identity $\displaystyle \left(x_{1}+x_{2}+x_{3}\right)^2 = x^2_{1}+x^2_{2}+x^2_{3}+2(x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1})$

So $\displaystyle x^2_{1}+x^2_{2}+x^2_{3} = 9^2-2(11) = 81-22 = 59$

 

[anemone]

Spoiler

Let $\tan^2 10^\circ = x_{1}$ and $\tan^2 50^\circ =x_{2}$ and $\tan^2 70^\circ = x_{3}$

Then $\displaystyle \tan^2(3\cdot 10^\circ ) = \frac{1}{3}$ and $\displaystyle \tan^2(3\cdot 50^\circ ) = \frac{1}{3}$ and $\displaystyle \tan^2(3\cdot 70^\circ ) = \frac{1}{3}$

Using $\displaystyle \tan (3x) = \frac{3\tan x-\tan^3 x}{1-3\tan^2 x} = \frac{1}{\sqrt{3}}$

So $\displaystyle \frac{1}{3} = \left(\frac{3\tan x-\tan^3 x}{1-3\tan^2 x}\right)^2\Rightarrow (1-3\tan^2 x)^2 = 3(3\tan x-\tan^3 x)^2$

So $\displaystyle 1+9\tan^4 x-6\tan^2 x = 27\tan^2 x+3\tan^6 x-18\tan^4 x$

So $\displaystyle 3\tan^6 x-27\tan^4 x+33\tan^2 x-1=0$

has a roots $x_{i}=\tan^2(\theta)\;,$ Where $\theta = 10^\circ\;,50^\circ\;,70^\circ$

So we can say $x_{1}\;,x_{2}\;,x_{3}$ are the roots of $3x^3-27x^2+33x-1=0$

So $\displaystyle x_{1}+x_{2}+x_{3} = 9$ and $x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1} = 11$

So Using Identity $\displaystyle \left(x_{1}+x_{2}+x_{3}\right)^2 = x^2_{1}+x^2_{2}+x^2_{3}+2(x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1})$

So $\displaystyle x^2_{1}+x^2_{2}+x^2_{3} = 9^2-2(11) = 81-22 = 59$

Bravo, jacks!(Cool)