A geometry problem by Aakhyat Singh

If sinx=cosy, then x = pi/2 - y, or y = -pi/2 + x.

Solving for the first case, we have picosA = pi/2 - pisinA, or (dividing by pi and moving sinA to the other side) cosA + sinA = 1/2. Squaring both sides, we have 1 + sin2A = 1/4, or sin2A = -3/4.

Solving the second case, we have pisinA = -pi/2 + picosA, or (simplifying like above) sinA - cosA = 1/2. Squaring both sides, we have 1 - sin2A = 1/4, or sin2A = 3/4.

Knowing that $cos(x) = sin(\frac{\pi}{2} - x)$ , we can write the above trig equation as:

$sin(\pi cos(A)) = cos(\pi sin(A)) = sin(\frac{\pi}{2} - \pi sin(A))$ ;

or $\pi cos(A) = \frac{\pi}{2} - \pi sin(A)$ ;

or $\sqrt{1 - sin^{2}(A)} = \frac{1}{2} - sin(A)$ ;

or $1- sin^{2}(A) = \frac{1}{4} - sin(A) + sin^{2}(A)$ ;

or $0 = -\frac{3}{4} - sin(A) + 2sin^{2}(A)$ ;

or $sin(A) = \frac{1 \pm \sqrt{1 - 4(2)(-\frac{3}{4})}}{4} = \frac{1 \pm \sqrt{7}}{4} \Rightarrow A \approx 65.705, -24.295$ degrees. We finally compute $sin(2A)$ according to $sin(2 \cdot 65.705) = \boxed{\frac{3}{4}}$ and $sin(2 \cdot -24.295) = \boxed{-\frac{3}{4}}.$