Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator

Question

The question is to find the exact value of:

$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$

without using a calculator.

I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $\tan{\left(\frac{2\pi}{5}\right)}$ to find that the answer is $10$. However, I want to know whether there is a faster way that does not involve calculating those values.

So far, I have this: let $a=\tan{\left(\frac{\pi}{5}\right)}$ and $b=\tan{\left(\frac{2\pi}{5}\right)}$; then, $b=\frac{2a}{1-a^2}$ and $a=-\frac{2b}{1-b^2}$, so multiplying the two and simplifying gives:

$$a^2+b^2={\left(ab\right)}^2+5$$

Any ideas? Thanks!

Answer 1

Let $a=\tan(\frac{\pi}{5})$ and $b= \tan(\frac{2 \pi}{5})$. Also let $s=a^2+b^2$ and $p=(a b)^2$. We have \begin{eqnarray*} a^2+b^2 &=&(ab)^2+5 \\ s&=&p+5. \end{eqnarray*} Now $a(1-b^2)=-2b$ and $b(1-a^2)=2a$ square these equations and add them together \begin{eqnarray*} 3(a^2+b^2)+4(ab)^2&=&(ab)^2(a^2+b^2) \\ 3s+4p&=&sp. \end{eqnarray*} Now eliminate $p$ and we have the quadratic $s^2-12s-20=0$. This has roots $2$ and $\color{red}{10}$.

Answer 2

Let $\tan\left(\frac{\pi}{5}\right)=x$

$$\tan\left(\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}\right)=0$$

$$\implies S_1-S_3+S_5=0$$

($S_k$ represents sum of tangents taken $k$ at a time)

$$\implies 5x-10x^3+x^5=0$$

Now the roots of this equation are $\tan\left(\frac{\pi}{5}\right),\tan\left(\frac{2\pi}{5}\right),\tan\left(\frac{3\pi}{5}\right),\tan\left(\frac{4\pi}{5}\right),\tan\left(\frac{\pi}{5}\right)$. Now sum of squares of the roots is $$\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)+\tan^2\left(\frac{3\pi}{5}\right)+\tan^2\left(\frac{4\pi}{5}\right)+\tan^2\left(\frac{\pi}{5}\right)=2\left(\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)\right)$$ Now this when calculated from basic concept of theory of equations applying to given polynomial comes to be $20$.

Hence $$2\left(\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)\right)=20 \implies \tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)=10$$

Answer 3

Given this article here, your problem would come out to $2\cdot4+2=10$. You should check the article out because it gives some links to other pages and possibly proofs.

My question here has a proof of the identity.

Answer 4

Inspired/ Trying to make sense of Jaideep's Answer ... \begin{eqnarray*} \tan(5 \theta) =\frac{5 \tan(\theta)-10 \tan^3(\theta)+\tan^5(\theta)}{1-10\tan^2(\theta) +5\tan^4(\theta)} \end{eqnarray*} Let $\theta=\frac{\pi}{5}$ and $x=tan(\frac{\pi}{5})$. We have the quintic \begin{eqnarray*} 5x-10x^3+x^5=0 \end{eqnarray*} This has roots $tan(\frac{\pi}{5}),tan(\frac{2\pi}{5}),tan(\frac{3\pi}{5}),tan(\frac{4\pi}{5}),tan(\frac{5\pi}{5})$.

Note that $tan(\frac{3\pi}{5})=-tan(\frac{2\pi}{5})$ , $tan(\frac{4\pi}{5})=-tan(\frac{\pi}{5}) $ and $tan(\frac{5\pi}{5})=0$.

Now square this quintic & substitute $y=x^2$ and we have \begin{eqnarray*} 25y-100y^2+110y^3-\color{red}{20}y^4+y^5=0 \end{eqnarray*} The sum of the roots of this polynomial gives $2 (tan^2(\frac{\pi}{5})+tan^2(\frac{2\pi}{5}))=20$.

Answer 5

Computing Tangents

Using that $1+\color{#090}{\cos\left(\frac{2\pi}5\right)}+\color{#C00}{\cos\left(\frac{4\pi}5\right)}+\color{#C00}{\cos\left(\frac{6\pi}5\right)}+\color{#090}{\cos\left(\frac{8\pi}5\right)}=0$, $$ \begin{align} 0 &=\color{#090}{\cos\left(\frac{2\pi}5\right)}\color{#C00}{+\cos\left(\frac{4\pi}5\right)}+\frac12\\ &=\color{#090}{2\cos^2\left(\frac{\pi}5\right)-1}\color{#C00}{-\cos\left(\frac{\pi}5\right)}+\frac12\\[3pt] &=4\cos^2\left(\frac{\pi}5\right)-2\cos\left(\frac{\pi}5\right)-1 \end{align} $$ Thus, $$ \cos\left(\frac{\pi}5\right)=\frac{1+\sqrt5}{4}\quad\text{and}\quad\cos\left(\frac{2\pi}5\right)=\frac{-1+\sqrt5}{4} $$ and therefore, $$ \sec\left(\frac{\pi}5\right)=\sqrt5-1\quad\text{and}\quad\sec\left(\frac{2\pi}5\right)=\sqrt5+1 $$ so that $$ \tan^2\left(\frac{\pi}5\right)=5-2\sqrt5\quad\text{and}\quad\tan^2\left(\frac{2\pi}5\right)=5+2\sqrt5 $$ So finally, $$ \bbox[5px,border:2px solid #C0A000]{\tan^2\left(\frac{\pi}5\right)+\tan^2\left(\frac{2\pi}5\right)=10} $$

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Using Residues (not for algebra-precalculus)

$\frac{5/z}{z^5-1}$ has residue $1$ at each of the fifth roots of unity and residue $-5$ at the origin.

The integral of $$ -\left(\frac{z^2-1}{z^2+1}\right)^2\frac{5/z}{z^5-1} $$ around a circle of very large radius vanishes since its absolute value $\sim\!\frac1{|z|^6}$. Thus, the sum of its residues must be $0$. If $|z|=1$, $-\left(\frac{z^2-1}{z^2+1}\right)^2=\tan^2(\arg(z))$. Thus, the sum of the residues at the fifth roots of unity is twice the quantity we want. This is the negative of the sum of the residues at $0$, $i$, and $-i$.

$\newcommand{\Res}{\operatorname*{Res}}$ $$ \begin{align} \Res_{z=i}\left(-\left(\frac{z^2-1}{z^2+1}\right)^2\frac{5/z}{z^5-1}\right)&=-\frac{25}2\\ \Res_{z=-i}\left(-\left(\frac{z^2-1}{z^2+1}\right)^2\frac{5/z}{z^5-1}\right)&=-\frac{25}2\\ \Res_{z=0}\left(-\left(\frac{z^2-1}{z^2+1}\right)^2\frac{5/z}{z^5-1}\right)&=5\\ \end{align} $$ Thus, the sum of the residues is $-\frac{25}2-\frac{25}2+5=-20$. Therefore, $$ 2\left(\tan^2\left(\frac{\pi}5\right)+\tan^2\left(\frac{2\pi}5\right)\right)=\sum_{k=0}^4\tan^2\left(\frac{2k\pi}5\right)=20 $$ And so $$ \bbox[5px,border:2px solid #C0A000]{\tan^2\left(\frac{\pi}5\right)+\tan^2\left(\frac{2\pi}5\right)=10} $$

Answer 6

$$\tan^2\frac{\pi}{5}+\tan^2\frac{2\pi}{5}=\frac{1-\cos\frac{2\pi}{5}}{1+\cos\frac{2\pi}{5}}+\frac{1-\cos\frac{4\pi}{5}}{1+\cos\frac{4\pi}{5}}=$$ $$=\frac{2-2\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}{1+\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}=$$ $$=\frac{2-\frac{4\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}{2\sin\frac{2\pi}{5}}}{1+\frac{2\sin\frac{\pi}{5}\cos\frac{2\pi}{5}+2\sin\frac{\pi}{5}\cos\frac{4\pi}{5}}{2\sin\frac{\pi}{5}}+\frac{4\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}{4\sin\frac{2\pi}{5}}}=$$ $$=\frac{2-\frac{\sin\frac{8\pi}{5}}{2\sin\frac{2\pi}{5}}}{1+\frac{\sin\frac{3\pi}{5}-\sin\frac{\pi}{5}+\sin\frac{5\pi}{5}-\sin\frac{3\pi}{5}}{2\sin\frac{\pi}{5}}+\frac{\sin\frac{8\pi}{5}}{4\sin\frac{2\pi}{5}}}=\frac{2+\frac{1}{2}}{1-\frac{1}{2}-\frac{1}{4}}=10.$$