Trigonometry with multiple angle and exact value of $\tan\pi/5$ By considering the equation $\tan5\theta=0$, show that the exact value of $\tan\pi/5$ is $\sqrt{5-2\sqrt{5}}$. 
Do I need to evaluate the multiple angle for $\tan5\theta=0$?

Using Euler's Formula and Binomial Theorem, we get
$$
\begin{align}
\cos(5\theta)+i\sin(5\theta)
&=\left(\cos(\theta)+i\sin(\theta)\right)^5\\
&=\cos^5(\theta)+5i\cos^4(\theta)\sin(\theta)-10\cos^3(\theta)\sin^2(\theta)\\
&-10i\cos^2(\theta)\sin^3(\theta)+5\cos(\theta)\sin^4(\theta)+i\sin^5(\theta)\tag{1}
\end{align}
$$
Taking the ratio of the real and imaginary parts of $(1)$, we get
$$
\tan(5\theta)=\frac{5\tan(\theta)-10\tan^3(\theta)+\tan^5(\theta)}{1-10\tan^2(\theta)+5\tan^4(\theta)}\tag{2}
$$
Thus, if $\tan(5\theta)=0$, but $\tan(\theta)\ne0$, $(2)$ says that
$$
5-10\tan^2(\theta)+\tan^4(\theta)=0\tag{3}
$$
The Quadratic Formula yields
$$
\tan^2(\theta)=5\pm2\sqrt{5}\tag{4}
$$
Therefore,
$$
\tan(\theta)=\pm\sqrt{5\pm2\sqrt{5}}\tag{5}
$$
Matching up the least positive values of $\theta$ for which $\tan(5\theta)=0$ yields
$$
\begin{align}
\tan\left(\frac\pi5\right)&=+\sqrt{5-2\sqrt{5}}\\
\tan\left(\frac{2\pi}5\right)&=+\sqrt{5+2\sqrt{5}}\\
\tan\left(\frac{3\pi}5\right)&=-\sqrt{5+2\sqrt{5}}\\
\tan\left(\frac{4\pi}5\right)&=-\sqrt{5-2\sqrt{5}}\\
\end{align}\tag{6}
$$
Note that these values support this result.
