Solving $\cos(2\theta)=\sin\left(\frac{\theta}{2}\right)$ Consider an acute angle $\theta$, which has the property that
$$\cos(2\theta)=\sin\left(\frac{\theta}{2}\right).$$
I am trying to find the value of $\theta$. Using the identity
$$\cos(2\theta)=\sin\left(\frac{\pi}{2}-2\theta\right),$$
I can write the first equation as
$$\sin\left(\frac{\pi}{2}-2\theta\right)=\sin\left(\frac{\theta}{2}\right).$$
Equating the argument of the $\sin$ functions, we see that
\begin{align}
\frac{\pi}{2}-2\theta&=\frac{\theta}{2} \\
\frac{5}{2}\theta&=\frac{\pi}{2} \\
\theta&=\frac{\pi}{5}.
\end{align}
While this yields the correct answer, I do not know how we can justify equating the arguments of the $\sin$ functions.

In general we have
$$\cos x = \sin y \iff y = \frac \pi 2 \pm x + 2k\pi$$
that is in this case
$$\frac \theta 2 = \frac \pi 2 \pm 2\theta  + 2k\pi$$
which leads to the general solutions

*

*$\theta_1 = \frac \pi 5+\frac 4 5 k \pi$

*$\theta_2 = -\frac \pi 3+\frac 4 3 k \pi$
and the solution for an acute angle is indeed $\theta = \frac \pi 5$.
