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 $\tan5\theta=0$?
 A: $$\tan(5x) = \dfrac{\tan(3x) + \tan(2x)}{1-\tan(3x) \tan(2x)} = \dfrac{\dfrac{3 \tan(x) - \tan^3(x)}{1-3 \tan^2(x)} + \dfrac{2 \tan(x)}{1-\tan^2(x)}}{1- \dfrac{3 \tan(x) - \tan^3(x)}{1-3 \tan^2(x)} \cdot \dfrac{2 \tan(x)}{1-\tan^2(x)}}$$
Hence,
$$\tan(5x) = 0 \implies (3t-t^3)(1-t^2) + 2t(1-3t^2) = t \left(t^4-4t^2+3 - 6t^2+2 \right) = 0$$
where $t= \tan(x)$. Since we are interested in $\tan(\pi/5)$, we can rule out $t=0$. Hence, we need to solve a bi-quadratic $t^4-10t^2+5 = 0$. This gives us
$$t^2 = \dfrac{10 \pm \sqrt{100-4 \times 5}}{2} = 5 \pm 2 \sqrt5 \implies t = \pm \sqrt{5 \pm 2\sqrt5}$$ We also know that
$\tan(\pi/5) \in (\tan(0),\tan(\pi/4)) = \left(0,1 \right)$. Hence, $$\tan(\pi/5) = \sqrt{5 - 2\sqrt5}$$
A: The idea is to expand $\tan 5\theta$ so as to get a function of $\tan \theta$. Then by letting $\theta = \pi/5$, you will get an equation in $\tan \pi/5$.
In details, put $x = \tan \pi/5$ $$ \tan 5 \theta = \frac{\tan 2\theta + \tan 3\theta}{1 - \tan 2\theta \tan 3\theta}$$
Then you find that $\tan 2\pi/5 + \tan 3\pi/5 = 0$.
But $$\tan 2\pi/5 = \frac{2 x}{1 - x^2}$$
And $$ \tan 3\pi/5 = \frac{\tan(2\pi/5) + x}{1 - x\tan(2\pi/5)} = \frac{3x - x^3}{1 - 3x^2}.$$
Conclusion $$ \frac{2x}{1 -x^2} + \frac{3x - x^3}{1 - 3x^2} = 0.$$
Because $x \neq 0$, we see that $x$ is solution of the polynomial $x^4 - 10x^2 +5 = 0$.
A: 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.
