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 $\tan5\theta=0$?

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 $x^4 - 10x^2 +5 = 0$.
