computing the value of $\sin^2(\frac{\pi}{10}$) The question asks me to prove the following formula: $$\cos5\theta=\cos\theta(16\sin^4\theta-12\sin^2\theta+1)$$ which is pretty straightforward to do using De Moivre's theorem. They further ask me to show that the exact value for $\sin^2\frac{\pi}{10}=\frac{3-\sqrt{5}}{8}$.
I proceed by simply solving the quadratic equation $$16\sin^4\frac{\pi}{10}-12\sin^2\frac{\pi}{10}+1=0$$ However, this yields two answers, namely $\frac{3\pm \sqrt{5}}{8}$. I don't know how to eliminate the answer where it's a plus and the justification from the book is

Since $\sin^2(\frac{\pi}{10})\lt \sin^2(\frac{3\pi}{10})$

I don't understand how that relates to the problem.
 A: The given identity states $$16 \sin^4 \theta - 12 \sin^2 \theta + 1 = \frac{\cos 5\theta}{\cos \theta}.$$  By substituting $\theta = \pi/10$, we get the first quadratic (i.e., $\sin^2 \theta$ is a root of the quadratic $16z^2 - 12z + 1$), since the RHS is zero as a result of $5\theta = \pi/2$.  But we can also note that $\theta = 3\pi/10$ implies $$\cos 5\theta = \cos \frac{3\pi}{2} = 0,$$ so the same quadratic has $\sin^2 \frac{3\pi}{10}$ as a root.  Now it becomes a simple matter to decide which root is which.
A: Say you were asked to find $\sin^2\left( \frac{3\pi}{10} \right) $.
You would proceed by saying that $$\cos \left( \frac{15\pi}{10} \right) = \cos \left( \frac{3\pi}{2} \right) = 0 \\
16\sin^4\frac{3\pi}{10}-12\sin^2\frac{3\pi}{10}+1=0$$
which is exactly the same quadratic equation you have solved when finding 
$\sin \left( \frac{3\pi}{10} \right)$.  Since you know that 
$\sin \left( \frac{\pi}{10} \right) < \sin \left( \frac{3\pi}{10} \right)$ you know that the smaller root of the quadratic is $\sin \left( \frac{\pi}{10} \right)$.
And by the way, you can actually find the square root of that:
$$\sin \left( \frac{\pi}{10} \right) = \sqrt\frac{{3-\sqrt{5}}}{8}
= \frac{\sqrt{5}-1}{4}$$
