12
$\begingroup$

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

Can anyone help me with this? Thank You!

$\endgroup$
13
$\begingroup$

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 $$ x^3-\frac{3}{4}x+\frac{1}{8}=0 $$

]

$\endgroup$
  • 1
    $\begingroup$ Method 1 is brilliant +1 $\endgroup$ – Shailesh Jul 18 '16 at 10:17
  • $\begingroup$ @ lab bhattacharjee, Could You explain me the method 2 of your answer? $\endgroup$ – pi-π Feb 10 '17 at 5:20
9
$\begingroup$

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 $$ 36-3=33 $$

$\endgroup$
5
$\begingroup$

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,

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

$\endgroup$
5
$\begingroup$

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 $$\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ) = 33$$

$\endgroup$
1
$\begingroup$

Since $$1+\tan^2 \alpha=\frac 1{\cos^2\alpha}=\frac 2{1+\cos 2\alpha},$$

$$\tan ^{2}20^{\circ}+\tan ^{2}40^{\circ}+\tan ^{2}80^{\circ}+3= \frac 2{1+\cos 40^{\circ}}+\frac 2{1+\cos 80^{\circ}}+\frac 2{1+\cos 160^{\circ}}.$$

Reducing the last sum to a common denominator and using the equalities from the appendix of the answer by lab bhattacharjee, we obtain $36$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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