## Saturday, May 16, 2015

### Third Method in Proving $\tan^2 20^{\circ}+\tan^2 40^{\circ}+\tan^2 80^{\circ}=33$

As a good mathematics educator, we should solve the best quality math problems using as many ways as we could, because each solution has its own learning value. Who knows, students might gain something from the long and tedious method of problem solving method and one day, they will surprise us with the improved version of the solution!

One thing every mathematics educator has to remember is that our imaginations are vibrant, our hearts are open, everything about math amazes us, and we think anything is possible.

No more shilly-shallying, I will start right away to teach you how to prove $\tan^2 20^{\circ}+\tan^2 40^{\circ}+\tan^2 80^{\circ}=33$ without using the trigonometric identity $\tan^2 x+\tan^2(60^{\circ}-x)+\tan^2(60^{\circ}+x)=9\tan^2 3x+6$.

Well, the method that I'm going to teach here, it isn't a very handsome method, but it makes plenty of sense and what matters most to me is that it teaches us from all angles related to other trigonometric areas.

If we let $\tan 40^{\circ}=x$, then we have:

$\tan 20^{\circ}=\tan (60^{\circ}-40^{\circ})=\dfrac{\sqrt{3}-x}{1+\sqrt{3}x}$ and

$\tan 80^{\circ}=\dfrac{2\tan 40^{\circ}}{1-\tan^2 40^{\circ}}=\dfrac{2x}{1-x^2}$

Subsequently, we obtain:

$\cos 20^{\circ}=\dfrac{1+\sqrt{3}x}{2\sqrt{1+x^2}}\implies \sec 20^{\circ}=\dfrac{2\sqrt{1+x^2}}{1+\sqrt{3}x}$;

$\cos 40^{\circ}=\dfrac{1}{\sqrt{1+x^2}}\implies \sec 40^{\circ}=\sqrt{1+x^2}$;

$\cos 80^{\circ}=\dfrac{1-x^2}{1+x^2}\implies \sec 80^{\circ}=\dfrac{1+x^2}{1-x^2}$;

Also, from $\tan 40^{\circ}=x$, we know that

$-\sqrt{3}=\tan 120^{\circ}=\tan 3(40^{\circ})=\dfrac{3\tan 40^{\circ}-\tan^3 40^{\circ}}{1-3\tan^2 40^{\circ}}=\dfrac{3x-x^3}{1-3x^2}$

We get a cubic equation in $x$ and we rewrite it to make $x^3$ the subject:

$x^3=\sqrt{3}+3x-3\sqrt{3}x^2$

And from there we get another expression of $x^4$ as a quadratic equation in $x$:

$x^4=\sqrt{3}x+3x^2-3\sqrt{3}x^3$

$\,\,\,\,\,\,\,\,=\sqrt{3}x+3x^2-3\sqrt{3}(\sqrt{3}+3x-3\sqrt{3}x^2)$

$\,\,\,\,\,\,\,\,=30x^2-8\sqrt{3}x-9$

Therefore, our main plan of attack is to reduce any $x$ term that is higher than $2$ down by using the two formulas above, so we will only deal with the quadratic expressions:

$\tan^2 20^{\circ}+\tan^2 40^{\circ}+\tan^2 80^{\circ}$

$\,\,\,=(\sec^2 20^{\circ}-1)+(\sec^2 40^{\circ}-1)+(\sec^2 80^{\circ}-1)$

$\,\,\,=(\sec^2 20^{\circ}\sec^2 40^{\circ}+\sec^2 80^{\circ}-3$

$\,\,\,=\sec^2 20^{\circ}+\sec^2 40^{\circ}+\sec^2 80^{\circ}-3$

$\,\,\,=\left(\dfrac{2\sqrt{1+x^2}}{1+\sqrt{3}x}\right)^2 +\left(\sqrt{1+x^2} \right)^2+\left(\dfrac{1+x^2}{1-x^2}\right)^2-3$

$\,\,\,=\dfrac{4(1+x^2)}{(1+\sqrt{3}x)^2} +1+x^2+\dfrac{(1+x^2)^2}{(1-x^2)^2}-3$

$\,\,\,=\dfrac{4(1+x^2)(1-x^2)^2+(1+x^2)^2(1+\sqrt{3}x)^2}{(1+\sqrt{3}x)^2(1-x^2)^2} +x^2-2$

$\,\,\,=\dfrac{4(932x^2-256\sqrt{3}x-284)+8(310x^2-84\sqrt{3}x-94)}{2188x^2-600\sqrt{3}x-668} +x^2-2$

$\,\,\,=\dfrac{6208x^2-1696\sqrt{3}x-1888}{2188x^2-600\sqrt{3}x-668} +x^2-2$

$\,\,\,=\dfrac{6208x^2-1696\sqrt{3}x-1888}{2188x^2-600\sqrt{3}x-668} +x^2-2$

$\,\,\,=\dfrac{6208x^2-1696\sqrt{3}x-1888+x^2(2188x^2-600\sqrt{3}x-668)}{2188x^2-600\sqrt{3}x-668}-2$

$\,\,\,=\dfrac{6208x^2-1696\sqrt{3}x-1888+70372x^2-19304\sqrt{3}x-21492}{2188x^2-600\sqrt{3}x-668}-2$

$\,\,\,=\dfrac{76580x^2-21000\sqrt{3}x-23380}{2188x^2-600\sqrt{3}x-668}-2$

$\,\,\,=\dfrac{35(2188x^2-600\sqrt{3}x-668)}{2188x^2-600\sqrt{3}x-668}-2$

$\,\,\,=\dfrac{35\cancel{(2188x^2-600\sqrt{3}x-668)}}{\cancel{(2188x^2-600\sqrt{3}x-668)}}-2$

$\,\,\,=35-2$

$\,\,\,=33$

and we're done.