how to calculate the exact value of $\tan \frac{\pi}{10}$ 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
 A: 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.
A: 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 $\tan \frac{\pi}{10}$.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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$$
\cos\pars{\bracks{n + 1}\theta} + \cos\pars{\bracks{n - 1}\theta}
=
2\cos\pars{n\theta}\cos\pars{\theta}
$$
$$
\cos\pars{\bracks{n + 1}\theta}
=
2\cos\pars{n\theta}\cos\pars{\theta}
-
\cos\pars{\bracks{n - 1}\theta}
$$
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}
$${\large%
\tan\pars{\pi \over 10} = \sqrt{1 - {2 \over \sqrt{5\,}}}}
$$
A: 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).
A: $$\tan\frac{3\pi}{10}=\tan(\frac{\pi}{2}-\frac{2\pi}{10})=\cot\frac{2\pi}{10}$$
$$\frac{3\tan\frac{\pi}{10}-\tan^3\frac{\pi}{10}}{1-3\tan^2\frac{\pi}{10}}=\frac{\cot^2\frac{\pi}{10}-1}{2\cot\frac{\pi}{10}}$$
$$(3\tan\frac{\pi}{10}-\tan^3\frac{\pi}{10})(2\cot\frac{\pi}{10})=(\cot^2\frac{\pi}{10}-1)(1-3\tan^2\frac{\pi}{10})$$
$$6-2\tan^2\frac{\pi}{10}=\cot^2\frac{\pi}{10}-4+3\tan^2\frac{\pi}{10}$$
$$5\tan^2\frac{\pi}{10}-10+\cot^2\frac{\pi}{10}=0$$
$$5\tan^4\frac{\pi}{10}-10\tan^2\frac{\pi}{10}+1=0$$
$$\tan^2\frac{\pi}{10}=\frac{10\pm\sqrt{100-20}}{10}=\frac{10\pm4\sqrt{5}}{10}=1+\frac{2}{\sqrt{5}}\;\textrm{or}\;1-\frac{2}{\sqrt{5}}(\textrm{rej.})$$
$$\tan\frac{\pi}{10}=\sqrt{1+\frac{2}{\sqrt{5}}}\;\textrm{or}\;-\sqrt{1+\frac{2}{\sqrt{5}}}(\textrm{rej.})$$
$$∴\tan\frac{\pi}{10}=\sqrt{1+\frac{2}{\sqrt{5}}}$$
A: 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 $0<\theta<45^\circ$.
