If $\sin(\pi \cos\theta) = \cos(\pi\sin\theta)$, then show ........ If $\sin(\pi\cos\theta) = \cos(\pi\sin\theta)$,
then show that $\sin2\theta = \pm 3/4$.
I can do it simply by equating $\pi - \pi\cos\theta$ to $\pi\sin\theta$,
but that would be technically wrong as those angles could be in different quadrants. So how to solve?
 A: $$\cos(\pi\sin\theta)=\sin(\pi\cos\theta)=\cos\left(\dfrac\pi2-\pi\cos\theta\right)$$
$$\implies\pi\sin\theta=2m\pi\pm\left(\dfrac\pi2-\pi\cos\theta\right)$$  where $m$ is any integer
$$\iff\sin\theta=2m\pm\left(\dfrac12-\cos\theta\right)$$
$$\iff\sin\theta\mp\cos\theta=2m\pm\dfrac12$$
$\sin\theta\mp\cos\theta=\sqrt2\sin\left(\theta\mp\dfrac\pi4\right)$
$\implies-\sqrt2\le\sin\theta\mp\cos\theta\le\sqrt2\implies m=?$
Now square both sides
A: We first rewrite
$$
\cos(\pi/2 - \pi \cos \theta) = \cos(\pi \sin \theta)
$$
(cofunction identity).
We notice that $\cos(x)$ is a periodic function with period $2\pi$, so we need a period offset term to be sure that we find all solutions. We also need to account for the fact that $\cos(x)$ is symmetric, so:
$$
\cos(\pi/2 - \pi \cos \theta) = \cos(\pm \pi \sin \theta + 2 \pi k) \\
\pi/2 - \pi \cos \theta = \pm \pi \sin \theta + 2 \pi k
$$
Then we do basic algebra
$$
1/2 - \cos \theta = \pm \sin \theta + 2k \\
\cos \theta \pm \sin \theta = 1/2 - 2k
$$
We'll get back to this.
Trigonometric trick time!
We can do a neat little trick to all functions of the form $a \cos \theta + b \sin \theta$. We rewrite:
\begin{align*}
a \cos \theta + b \sin \theta &= \sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}} \cos \theta + \frac{b}{\sqrt{a^2+b^2}} \sin \theta\right) \\ 
&=: \sqrt{a^2+b^2} (a' \cos \theta + b' \sin \theta) \\
&=: \sqrt{a^2+b^2} (\cos \phi \cos \theta + \sin \phi \sin \theta) \\
&=: \sqrt{a^2+b^2} \cos (\theta - \phi)
\end{align*}
where we can find $\phi=\arctan(b'/a') + \pi n = \arctan(b/a) +\pi n$ (note the $\pi n$ because, like $\sin(x)$ and $\cos(x)$, $\tan(x)$ is also a periodic function, so we have to account for all possible inverse values).
End trigonometric trick
We apply the trick to get
$$
\cos \theta \pm \sin \theta = 1/2 - 2k \\
\sqrt{1^2+1^2} \cos (\theta - \arctan (\pm1/1)) = 1/2 - 2k \\
\sqrt 2 \cos (\theta \mp \pi/4) = 1/2 - 2k
$$
(You can verify that the rewritten forms do indeed evaluate to the original.)
We continue:
$$
\cos (\theta \mp \pi/4) = 1/(2\sqrt{2}) + \sqrt{2} k = \sqrt{2} (1/4 - k)
$$
Notice that $-1 \le \cos(x) \le 1$, so only $k = 0$ is valid. Hence we get
$$
\cos (\theta \mp \pi/4) = \sqrt{2}/4
$$
Next, we notice
\begin{align*}
\sin 2\theta &= \cos (\pi/2 - 2 \theta) \\
&= \cos (2 \theta - \pi/2) \\
&= \cos(2 (\theta - \pi/4)) \\
&= 2 \cos^2 (\theta - \pi/4) - 1 \\
&= 2 (\sqrt{2}/4)^2 - 1 \\
&= 1/4 - 1 \\
&= -3/4
\end{align*}
We notice the other solution
\begin{align*}
\sin 2\theta &= -\cos (\pi/2 - 2 \theta) \\
&= \cos (2 \theta - \pi/2) \\
&= -\cos (2 \theta - \pi/2 + \pi) \\
&= -\cos (2 \theta + \pi/2) \\
&= -\cos(2 (\theta + \pi/4)) \\
&= -(2 \cos^2 (\theta + \pi/4) - 1) \\
&= -(2 (\sqrt{2}/4)^2 - 1) \\
&= -(1/4 - 1) \\
&= 3/4
\end{align*}
