Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Can anyone help me with this? Thank You!

share|improve this question
add comment

4 Answers

up vote 8 down vote accepted

Method $1:$

We know $$ \tan^2A=\frac{1-\cos2A}{1+\cos2A} $$

Let us find the cubic equation whose roots are $\cos40^\circ, \cos80^\circ, \cos160^\circ$.

As $\cos(3\cdot 40^{\circ})=\cos120^{\circ}=-\frac{1}{2}$ or, $4\cos^340^{\circ} -3\cos40^{\circ}=-\frac{1}{2}$.

So, $\cos40^{\circ} $ is a root of $$ 4x^3-3x=-\frac12\implies 8x^3-6x+1=0 $$ Similarly, $\cos80^{\circ},\cos160^{\circ}$ are also the roots of $ 8x^3-6x+1=0 $ (Another derivation can be found at the bottom)

If we replace $x$ with $\dfrac{1-y}{1+y}$, the sum of the roots of the new equation in $y$ will give us the desired value.

Method $2:$ (Inspired by Zarrax's answer)

Observe that $\tan(3\cdot20^\circ)=\tan60^\circ=\sqrt3$

$\tan(3\cdot40^\circ)=\tan120^\circ=\tan(180^\circ-60^\circ)=-\tan60^\circ=-\sqrt3$ $\iff \tan\{3(-40^\circ)\}=\sqrt3$

and $\tan(3\cdot80^\circ)=\tan240^\circ=\tan(180^\circ+60^\circ)=\tan60^\circ=\sqrt3$

$$\text{As }\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$$

$$\text{the roots of the equation } t^3-3\sqrt3t^2-3t+\sqrt3=0 (\text{ Putting } \tan3\theta=\sqrt3)$$ will be $\tan20^\circ,\tan(-40^\circ)=-\tan40^\circ, \tan80^\circ$

Using Vieta's formulas, $$\tan20^\circ+(-\tan40^\circ)+\tan80^\circ=\frac{3\sqrt3}1$$

$$\text{and } \tan20^\circ(-\tan40^\circ)+\tan20^\circ\cdot\tan80^\circ+\tan80^\circ(-\tan40^\circ)=-3$$

$$\text{So,}\tan^220^\circ+\tan^240^\circ+\tan^280^\circ =(\tan20^\circ)^2+(-\tan40^\circ)^2+(\tan80^\circ)^2$$ $$=\{\tan20^\circ+(-\tan40^\circ)+\tan80^\circ\}^2$$ $$-2\{\tan20^\circ(-\tan40^\circ)+\tan20^\circ\cdot\tan80^\circ+\tan80^\circ(-\tan40^\circ)\}$$

$$=(3\sqrt3)^2-2(-3)=33$$

[

Applying the following identities, $$\begin{align*} \cos 2A+\cos 2B&=2\cos(A-B)(A+B),\\ \sin2A&=2\sin A\cos A,\\ 2\cos A\cos B&=\cos(A-B)+\cos(A+B) \end{align*}$$
we get $$\begin{align*} \cos40^{\circ} + \cos80^{\circ} + \cos160^{\circ}&=0\\ \cos40^{\circ}\cos80^{\circ} + \cos80^{\circ}\cos160^{\circ} + \cos160^{\circ}\cos40^{\circ}&=-\frac{3}{4}\\ \end{align*}$$

$$\text{ and } \cos40^{\circ} \cos80^{\circ} \cos160^{\circ}=-\frac{1}{8}$$

Then the cubic equation whose roots are $\cos40^{\circ}, \cos80^{\circ}, \cos160^{\circ}$ is $$ x^3-\frac{3}{4}x+\frac{1}{8}=0 $$

]

share|improve this answer
add comment

In $(7)$ from this answer, it is shown that $$ \sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)=n(2n+1) $$ In this case, $n=4$ and you're missing $l=3$. $\tan^2(60^\circ)=3$, so the sum would be $$ 36-3=33 $$

share|improve this answer
    
It seems my solution is intimately related to yours... –  J. M. Jul 28 '12 at 11:27
add comment

Here's a linear algebraic route: from this answer, we find that the eigenvalues of the $4\times4$ min-matrix

$$\mathbf M=\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 \\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4 \end{pmatrix}$$

are $\lambda_k=\dfrac14\sec^2\left(\dfrac{k\pi}{9}\right)$ for $k=1,\dots,4$. From this, we have that the eigenvalues of $4\mathbf M-\mathbf I$ are $\nu_k=\tan^2\left(\dfrac{k\pi}{9}\right)$, and since the sum of the eigenvalues is equal to the trace of the matrix,

$$\tan^2\frac{\pi}{9}+\tan^2\frac{2\pi}{9}+\tan^2\frac{4\pi}{9}=4(1+2+3+4)-4-\tan^2\frac{\pi}{3}=33$$

share|improve this answer
add comment

Notice that for $\theta = 20, 40,$ and $80$ degrees you have $\tan^2(3\theta) = 3$. The tangent triple angle formula, which you can get from the tangent angle addition formula, says that $$\tan(3\theta) = {3\tan(\theta) - \tan^3(\theta) \over 1 - 3 \tan^2(\theta)}$$ So the equation $\tan^2(3\theta) = 3$ can be expressed as $$(3\tan(\theta) - \tan^3(\theta))^2 = 3(1 - 3 \tan^2(\theta))^2$$ After a little algebra, this becomes the following, where $x = \tan(\theta)$. $$x^6 - 33x^4 + 27x^2 - 3 = 0$$ By the above, this has roots $x = \tan(20^\circ), \tan(40^\circ),$ and $\tan(80^\circ)$. Since $x$ only appears to even powers here, the other roots must be $x = -\tan(20^\circ), -\tan(40^\circ),$ and $-\tan(80^\circ)$. The sum of the squares of all six roots is thus given by $2(\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ))$. However, if we write these roots as $r_1,...,r_6$, then we also have $$\sum_i r_i^2 = \left(\sum_i r_i\right)^2 - 2\sum_{i < j} r_ir_j$$ But $\sum_i r_i$ is the coefficient of $x^5$ in the above equation, namely zero, and $\sum_{i < j} r_ir_j$ is the coefficient of $x^4$, namely $-33$. So you get $$2(\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ)) = -2\times-33$$ So we conclude that $$\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ) = 33$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.