How to derive $\sin(\frac{\pi}{4}) = 2\sin(\frac{\pi}{8})(1-\sin^2(\frac{\pi}{8}))^\frac{1}{2}$

Question

Looking at a solution to the following problem:

If $s = \sin(\frac{\pi}{8})$, prove that $8^4 − 8s^2 + 1 = 0$

I see that the first step is using:

$$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$

to arrive at:

$$\sin\left(\frac \pi 4 \right) = 2s(1-s^2)^\frac 1 2$$

But I'm not sure how this is done. I've gotten this far:

$$\sin\left(\frac \pi 4 \right) = 2\sin\left(\frac \pi 8 \right) \cos\left(\frac \pi 8 \right) = 2s\cos\left(\frac \pi 8 \right)$$

But I'm not sure how $\cos(\frac{\pi}{8})$ relates to $(1-s^2)^\frac{1}{2}$.

If it's of any help, this is problem 1.10 from Riley's "Mathematical methods for physics and engineering".

Answer 1

hint: $\cos = \sqrt{1-\sin^2}$. The reason the absolute value is not there is because $\pi/8$ is an acute angle and there is no worry about this.

Answer 2

It's $$\sin\frac{\pi}{4}=2\sin\frac{\pi}{8}\cos\frac{\pi}{8},$$ which is obvious.