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.
 A: $$
\begin{aligned}
\operatorname{cos} 2 \theta &=\sin \frac{\theta}{2}=\cos \left(\frac{\pi}{2}-\frac{\theta}{2}\right) \\
2 \theta &=2 n \pi \pm\left(\frac{\pi}{2}-\frac{\theta}{2}\right) \\
\frac{5 \theta}{2} &=2 n \pi+\frac{\pi}{2} \text { or } \frac{3 \theta}{2}=2 n \pi-\frac{\pi}{2} \\
\theta &=\frac{(4 n+1) \pi}{5} \text { or } \frac{(4 n-1) \pi}{3},
\end{aligned}
$$
where $n\in Z.$
A: 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 $\theta = \frac \pi 5$.
A: Note that the solution to $$\sin(x) = \sin(y)$$ is $$x = 2k\pi + y \text{ and } x = 2k\pi + \pi - y$$ So in this case we have $$\frac\pi 2 - 2\theta = 2k\pi  + \frac\theta 2 \text{ and } \frac\pi 2 - 2\theta = 2k\pi  + \pi - \frac\theta 2$$The solutions are $$\frac{5\theta}{2} = \frac \pi 2 - 2k\pi \implies \theta = \frac \pi 5 - \frac{4k\pi}{5}, \ \ k \in \mathbb{Z} \\ \frac{3\theta}{2} = -\frac \pi 2 - 2k\pi \implies \theta = -\frac \pi 3 - \frac{4k\pi}{3}, \ \ k \in \mathbb{Z}$$
A: Like you said, you have to explain why $\sin x = \sin y \implies x = y$ in this case. This is called injectivity and, for real functions like sine, you have a neat geometrical reasoning: a function is injective when, whenever you draw a line from a point on the graph parallel to the $x$-axis, the line will not intercept another point on the graph.
Note that, on your specific case, $$\begin{align*}0 \le \theta < \frac\pi 2 &\implies 0 \le 2\theta < \pi\\ &\implies \frac\pi2\ge \frac\pi2 - 2\theta >  -\frac\pi2. \end{align*}$$
And, by the same kind of argument, $0 \le \frac\theta2 \le \frac\pi4$. Therefore both $\frac\theta2$ and $\frac\pi2-2\theta$ lie on the interval $(-\frac\pi2,\frac\pi2]$. Good news! The sine function is injective on this interval! We have the following graph:
                                                 
So we can, in fact, use the equality $\sin x = \sin y \implies x = y$, like you did. Yay!
Like other have said, in general there are more (and usually infinitely more) solutions to those equations, since $\sin x$ is $2\pi$-periodic. There is only one solution here only because $\theta$ is acute.

Remark: This is a naive argument; obviously you can make things harder by studying the general behaviour of the sine function or, if you like, prove that the sine is strictly increasing on this interval. Either way, the naive analysis still works.
A: That guarantees that that is one of the answers.
It's justified because if $x = y$ then $f(x) = f(y)$ for any possible function.
So IF $\frac \theta 2 = \frac \pi 2 - 2\theta$ then $\sin \frac \theta 2 = \sin (\frac \pi 2 -2\theta) = \cos 2\theta$, then any solution to $\frac \theta 2 = \frac \pi 2 - 2\theta$  (there is only one such solution) will be a solution to $\sin \frac \theta 2 = \cos 2\theta$.
But it doesn't mean it is the only solution.
It is true that if $x = y \implies f(x) = f(y)$ but it is not true that $f(x) = f(y) \implies x=y$.  $x=y$ will be one pair of solutions (always) but there may be others.
But we can take this further.
As $\cos 2\theta = \sin (\frac \pi 2 - 2\theta)$ always, it will suffice to find all solutions to $\sin\frac {\theta} 2 = \sin (\frac \pi 2 - 2\theta)$.
So we have to ask ourselves how do we find all solutions to $\sin x = \sin y$.
Well $x = y$ is one but so is $x = y + 2k\pi$ for any integer $k$.  And $x = \pi - y$ is another but so is $x = \pi- y +2j\pi=-y + (2j+1)\pi$ for any integer $j$.  And that is all of them.
So if $\sin\frac {\theta} 2= \sin (\frac \pi 2- 2\theta)$ then we have either

$\frac \theta 2 = \frac \pi 2 - 2\theta + 2k\pi$ for some integer $k$.

Or we have

$\frac \theta 2 = 2\theta - \frac \pi 2 + (2j+1)\pi$ for some integer $j$.

....
Now we just solve those.
If $\frac \theta 2 = \frac \pi 2 - 2\theta + 2k\pi$ then
$\frac 52\theta = \frac \pi 2 + 2k \pi$ so
$\theta = \frac \pi 5 + \frac 45k\pi$.
That is $\frac \pi 5$ is a solution but so are $\pi, \frac 95\pi, \frac{13}5\pi, \frac {17}5\pi$ etc.
And if $\frac \theta 2 = 2\theta - \frac \pi 2 + (2j+1)\pi$ then we have
$\frac 32\theta = \frac \pi 2 - (2j+1)\pi$.
(For simplicity sake we can replace $-(2j+1)$ with any other odd integer so lets replace it with a positive $(2m + 1)$ to get... )
$\frac 32\theta =\frac \pi 2 + (2m+1)\pi$
$\theta = \frac \pi 3 + \frac {4m+2}3\pi$.
So $\theta = -\frac{\pi} 3 \pi, \frac {7\pi}3, \frac {11\pi}3$ are all solutions.
