Having trouble understanding the steps of this half angle identity

Question

How does the solution go from $\sqrt{2\left(1+\cos\left(\frac{\pi}{2^{n+1}}\right)\right)}$ to $4\cos^2(\ldots)$? Where does the $4$ come from? I understand that the identity is $\cos^2(2x) = \cos(1+\cos(2x))/2$, but still lost on the $4$.

Answer 1

The half-angle identity states $$\cos^2 \frac{x}{2} = \frac{1+\cos x}{2},$$ which is the algebraic rearrangement of the double-angle identity $$\cos 2\theta = 2 \cos^2 \theta - 1.$$ Thus, with the choice $x = \pi/2^n$, we immediately obtain the desired transformation.

Answer 2

The simple fact is that

$$ 2 = 4 \cdot \frac12.$$

Therefore you can substitute for the first $2$ in $\sqrt{2\left(1+\cos\left(\frac{\pi}{2^{n+1}}\right)\right)}$ as follows:

$$\sqrt{2\left(1+\cos\left(\frac{\pi}{2^{n+1}}\right)\right)} = \sqrt{4 \cdot \frac12\left(1+\cos\left(\frac{\pi}{2^{n+1}}\right)\right)}.$$

And then you can apply the trig identity $\cos^2(2x) = \frac12(1+\cos(2x))$ to $\frac12\left(1+\cos\left(\frac{\pi}{2^{n+1}}\right)\right)$, using $\frac{\pi}{2^{n+1}}$ as $2x$.