Trigonometry with multiple angle and exact value of $\tan\pi/5$

Question

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 $\tan5\theta=0$?

Answer 1

$$\tan(5x) = \dfrac{\tan(3x) + \tan(2x)}{1-\tan(3x) \tan(2x)} = \dfrac{\dfrac{3 \tan(x) - \tan^3(x)}{1-3 \tan^2(x)} + \dfrac{2 \tan(x)}{1-\tan^2(x)}}{1- \dfrac{3 \tan(x) - \tan^3(x)}{1-3 \tan^2(x)} \cdot \dfrac{2 \tan(x)}{1-\tan^2(x)}}$$ 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, $$\tan(\pi/5) = \sqrt{5 - 2\sqrt5}$$

Answer 2

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

Answer 3

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.

Answer 4

The equation $\tan(5\theta)=0$ is rather clumsy to start with, really; but using the binary decomposition $5=4+1$ offers a solution that involves only quadratic equations.

Start with

$0=\tan(4\theta+\theta)=\dfrac{\tan(4\theta)+\tan(\theta)}{1-\tan(4\theta)\tan(\theta)}$

where the formula for the tangent of a sum is used.

The fraction is zero only if the numetator us, so $\tan(4\theta)=-\tan(\theta)$. We then apply the double angle formula for tangent twice:

$\tan(4\theta)=\dfrac{2\tan(2\theta)}{1-\tan^2(2\theta)}=\dfrac{4\tan(\theta)}{[1-\tan^2(\theta)]×[1-\dfrac{4\tan^2(\theta)}{[1-\tan^2(\theta)]^2}]}=-\tan(\theta)$

Then upon canceling a factor of $-\tan(\theta)$ this may be simplified to

$-4=[1-\tan^2(\theta)]×[1-\dfrac{4\tan^2(\theta)}{[1-\tan^2(\theta)]^2}]=1-\tan^2(\theta)-\dfrac{4\tan^2(\theta)}{1-\tan^2(\theta)}$

$\implies \tan^4(\theta)-10\tan^2(\theta)+5=0$

We solve this as a quadratic equation for $\tan^2(\theta)$ and take square roots. The four resulting roots correspond to $\tan(n\pi/5), n\in\{1,2,3,4\}$ as all these $\theta$ values give $\tan(5\theta)=0$. We choose the smallest positive root for $n=1$, thus $\tan(\pi/5)=\sqrt{5-2\sqrt5}$. The other positive root corresponds to $\tan(2\pi/5)=\sqrt{5+2\sqrt5}$.