Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$.
Can anyone help me with this? Thank You!
Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$.
Can anyone help me with this? Thank You!
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 $
]
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 $
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 $\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ) = 33$
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,
$\tan^2\frac{\pi}{9}+\tan^2\frac{2\pi}{9}+\tan^2\frac{4\pi}{9}=4(1+2+3+4)-4-\tan^2\frac{\pi}{3}=33$
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 $36$.