Thank you! I understood one crucial thing: we need to substitute in pie into LHS tan(5theta) so that we get theta=pie/5, which is one of our roots. And, since tan(pie) = 0, we can set our RHS expression = 0 too. By the same intuition, tan(2pie) is zero. Thus, overall, I understood the idea behind setting tan(5theta) = 0.
But what I don't get is why did you even think about tan(pie) = 0 in the first place -- then going on to using this to solve the problem. That is, you could have instead chosen, say, tan(pie/4) = 1, but you didn't. Why not? What gave you the idea to use tan(pie) = 0?
We previously derived the formula:
[MATH]\tan(5\theta)=\frac{t^5-10t^3+5t}{5t^4-10t^2+1}[/MATH]
This gives us a way to relate \(\tan(5\theta)\) to \(\tan(\theta)\). Now consider that:
[MATH]\pi=5\cdot\frac{\pi}{5}[/MATH]
[MATH]2\pi=5\cdot\frac{2\pi}{5}[/MATH]
And so if \(\theta=n\dfrac{\pi}{5}\) then \(5\theta=n\pi\).
The reason I chose values for which \(\tan(5\theta)=0\) is because we are searching for roots of a quadratic. Using the formula relating \(\tan(5\theta)\) to \(\tan(\theta)\), we see that in order for \(\tan(5\theta)=0\) we must have:
[MATH]\tan^5(\theta)-10\tan^3(\theta)+5\tan(\theta)=0[/MATH]
[MATH]\tan(\theta)\left(\tan^4(\theta)-10\tan^2(\theta)+5\right)=0[/MATH]
Now, when we have:
[MATH]0<\theta<\pi[/MATH]
We know \(\tan(\theta)\ne0\), and since both angles under consideration fall within that range, we may divide by \(\tan(\theta)\) to obtain the quadratic in \(\tan^2(\theta)\):
[MATH]\tan^4(\theta)-10\tan^2(\theta)+5=0[/MATH]
Since \(\tan(k\pi)=0\) where \(k\in\mathbb{Z}\), we know the above quadratic is true for all:
[MATH]\theta=k\frac{\pi}{5}[/MATH]
Because of the fact that on the unit circle, an angle of \(\theta\) is equivalent to \(\theta+2k\pi\), we know then that for:
[MATH]\theta\in\left\{k\frac{\pi}{5},(k+1)\frac{\pi}{5}\right\}[/MATH]
[MATH]\tan^2(\theta)[/MATH] must be distinct roots of the quadratic:
[MATH]x^2-10x+5=0[/MATH]
Now given that the product of the two roots of a quadratic is the quotient of the constant term divided by the coefficient of the squared term, we may conclude that:
[MATH]\tan^2\left(k\frac{\pi}{5}\right)\tan^2\left((k+1)\frac{\pi}{5}\right)=5[/MATH]
Since for \(k=1\) we have \(\theta\) in the first quadrant, we can therefore state:
[MATH]\tan\left(\frac{\pi}{5}\right)\tan\left(\frac{2\pi}{5}\right)=\sqrt{5}[/MATH]
Does this make sense?