how to calculate the exact value of $\tan \frac{\pi}{10}$

Question

I have an extra homework: to calculate the exact value of $ \tan \frac{\pi}{10}$. From WolframAlpha calculator I know that it's $\sqrt{1-\frac{2}{\sqrt{5}}} $, but i have no idea how to calculate that.

Thank you in advance, Greg

Answer 1

Your textbook probably has an example, where $\cos(\pi/5)$ (or $\sin(\pi/5)$) has been worked out. I betcha it also has formulas for $\sin(\alpha/2)$ and $\cos(\alpha/2)$ expressed in terms of $\cos\alpha$. Take it from there.

Answer 2

look this How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$?

then you will get $\sin \frac{\pi}{10}$($\frac{2\pi }{5}+\frac{\pi}{10}=\frac{\pi}{2}$) ,then $\tan \frac{\pi}{10}$.

Answer 3

$$\tan\frac{3\pi}{10}=\tan(\frac{\pi}{2}-\frac{2\pi}{10})=\cot\frac{2\pi}{10}$$

$$\frac{3\tan\frac{\pi}{10}-\tan^3\frac{\pi}{10}}{1-3\tan^2\frac{\pi}{10}}=\frac{\cot^2\frac{\pi}{10}-1}{2\cot\frac{\pi}{10}}$$

$$(3\tan\frac{\pi}{10}-\tan^3\frac{\pi}{10})(2\cot\frac{\pi}{10})=(\cot^2\frac{\pi}{10}-1)(1-3\tan^2\frac{\pi}{10})$$

$$6-2\tan^2\frac{\pi}{10}=\cot^2\frac{\pi}{10}-4+3\tan^2\frac{\pi}{10}$$

$$5\tan^2\frac{\pi}{10}-10+\cot^2\frac{\pi}{10}=0$$

$$5\tan^4\frac{\pi}{10}-10\tan^2\frac{\pi}{10}+1=0$$

$$\tan^2\frac{\pi}{10}=\frac{10\pm\sqrt{100-20}}{10}=\frac{10\pm4\sqrt{5}}{10}=1+\frac{2}{\sqrt{5}}\;\textrm{or}\;1-\frac{2}{\sqrt{5}}(\textrm{rej.})$$

$$\tan\frac{\pi}{10}=\sqrt{1+\frac{2}{\sqrt{5}}}\;\textrm{or}\;-\sqrt{1+\frac{2}{\sqrt{5}}}(\textrm{rej.})$$

$$∴\tan\frac{\pi}{10}=\sqrt{1+\frac{2}{\sqrt{5}}}$$

Answer 4

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $$ \cos\pars{\bracks{n + 1}\theta} + \cos\pars{\bracks{n - 1}\theta} = 2\cos\pars{n\theta}\cos\pars{\theta} $$ $$ \cos\pars{\bracks{n + 1}\theta} = 2\cos\pars{n\theta}\cos\pars{\theta} - \cos\pars{\bracks{n - 1}\theta} $$ Let's $\theta = \pi/10$ and $x = \cos\pars{\theta}$: \begin{align} \cos\pars{2\theta} &= 2x^{2} - 1 \tag{1} \\ \cos\pars{3\theta} &= \bracks{2\cos\pars{2\theta} - 1}x \tag{2} \\ \cos\pars{4\theta} &= 2\cos\pars{3\theta}x - \cos\pars{2\theta} \tag{3} \\ 0 &= 2\cos\pars{4\theta}x - \cos\pars{3\theta} \tag{4} \end{align}

$\pars{3}$ and $\pars{4}$ yield: $$ 0 = \pars{4x^{2} - 1}\cos\pars{3\theta} - 2\cos\pars{2\theta}x \tag{5} $$

$\pars{2}$ and $\pars{5}$ yield: $$ 0 = \pars{4x^{2} - 1}\bracks{2\cos\pars{2\theta} - 1}x - 2\cos\pars{2\theta}x = \pars{8x^{3} - 4x}\cos\pars{2\theta} - 4x^{3} + x $$ $$ 4\pars{2x^{2} - 1}\cos\pars{2\theta} - 4x^{2} + 1 = 0 \tag{6} $$

$\pars{1}$ and $\pars{6}$ yield: $$ 0 = 4\pars{2x^{2} - 1}^{2} - 4x^{2} + 1 = 4\pars{2x^{2} - 1}^{2} - 2\pars{2x^{2} - 1} - 1 $$

Then, $$ \pars{2x^{2} - 1}_{\pm} = {2 \pm \sqrt{\pars{-2}^{2} - 4\times 4\times\pars{-1}} \over 2\times 4} = {1 \pm \sqrt{5} \over 4} $$ Obviously, we take the "$+$ sign" as a solution: $$ 2x^{2} - 1 = {1 + \sqrt{5} \over 4} \quad\imp\quad x = \cos\pars{\pi \over 10} = \sqrt{{1 \over 2}\bracks{1 + {1 + \sqrt{5} \over 4}}} = \sqrt{{5 + \sqrt{5} \over 8}} $$ Then, \begin{align} \tan\pars{\pi \over 10} &= \sqrt{\bracks{1 \over \cos\pars{\pi/10}}^{2} - 1} = \sqrt{{8 \over 5 + \sqrt{5\,}} - 1} = \sqrt{3 - \sqrt{5\,} \over 5 + \sqrt{5\,}} = \sqrt{1 - 2\,{1 + \sqrt{5\,} \over 5 + \sqrt{5}}} \\[3mm]&= \sqrt{1 - 2\,{\pars{1 + \sqrt{5\,}}\pars{5 - \sqrt{5\,}} \over 20}} = \sqrt{1 - {4\sqrt{5\,} \over 10}} = \sqrt{1 - {2 \over \sqrt{5\,}}} \\[5mm]& \end{align}

$${\large% \tan\pars{\pi \over 10} = \sqrt{1 - {2 \over \sqrt{5\,}}}} $$

Answer 5

If $10x=\pi$ $\sin 2x=\cos 3x$ as $2x+3x=5x=\frac{\pi}{2}$ $\implies2\sin x \cos x=4\cos^3x-3\cos x$

$\implies 2\sin x=4\cos^2x-3$ as $\cos x≠0$

If $\sin x=t, 2t=4(1-t^2)-3\implies 4t^2+2t-1=0$

$$\implies t=\frac{-1±\sqrt{5}}{4}$$, but $\sin x>0$ as $0<x<\pi$

$$\sin \frac{\pi}{10}=\frac{\sqrt{5}-1}{4}$$

(1)So, $$\cos \frac{\pi}{10}=\sqrt{1-(\sin \frac{\pi}{10})^2}=\frac{\sqrt{10+2\sqrt5}}{4}$$

So, $$\tan \frac{\pi}{10}=\frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt5}}{4}}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt5}}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt5}}$$

$$=\sqrt{\frac{(\sqrt 5 -1)^2}{10+2\sqrt5}}=\sqrt{\frac{3-\sqrt 5}{\sqrt 5(\sqrt 5+1)}}=\sqrt{\frac{(3-\sqrt 5)(\sqrt 5 -1)}{\sqrt 5(\sqrt 5+1)(\sqrt 5 -1)}}=\sqrt{\frac{\sqrt 5-2}{\sqrt 5}}$$

Or(2) $$\cos \frac{\pi}{5}=1-2(\frac{\sqrt{5}-1}{4})^2=\frac{\sqrt 5 + 1}{4}$$

We know $$\cos2y=\frac{1-\tan^2y}{1+\tan^2y}\implies \tan^2y=\frac{1-\cos2y}{1+\cos2y}$$

So, $$\tan^2 \frac{\pi}{10}= \frac{1-\frac{\sqrt 5 + 1}{4}}{1+\frac{\sqrt 5 + 1}{4}}=\frac{3-\sqrt 5}{\sqrt 5(\sqrt 5+1)}$$ which we have already encountered in (1).

Answer 6

Let $\theta=\frac\pi{10}$ and $\tan\theta=x$. Then $5\theta=\frac\pi2$ so \begin{align} \tan4\theta&=\frac1x\tag{1} \end{align} By twice using the double-angle tan formula, \begin{align*} \tan2\theta&=\frac{2x}{1-x^2}\\ \tan4\theta&=\frac{2(\frac{2x}{1-x^2})}{1-(\frac{2x}{1-x^2})^2}\\ \frac1x&=\frac{4x(1-x^2)}{(1-x^2)^2-(2x)^2}\tag{by (1)}\\ (1-x^2)^2-4x^2&=x\times4x(1-x^2)\\ 1-2x^2+x^4-4x^2&=4x^2-4x^4\\ 5x^4-10x^2+1&=0\\ \implies x^2&=\frac{10\pm\sqrt{10^2-4\times5}}{2\times5}\\ &=1-\frac{\sqrt{5^2-5}}5\\ &=1-\frac{\sqrt{5-1}\sqrt5}5\\ &=1-\frac2{\sqrt5}\\ \implies x&=\sqrt{1-\frac2{\sqrt5}} \end{align*} signs being chosen because $0<x<1$ because $0<\theta<45^\circ$.