Find $\frac{b}{a}$ of trigonometry system of equation. Given
\begin{align}
\cos a + \cos b =0\\
\sin a +\sin b = \sqrt{2}
\end{align}
with constraint: $0<a<b<2\pi$. Determine $\frac{b}{a}$.
I try to change the system into
\begin{align}
\cos^2 a + \cos^2 b + 2\cos a\cos b=0\\
\sin^2 a +\sin^2 b + 2\sin a\sin b = 2
\end{align}
Adding two equation above become
\begin{align}
\cos(a-b)=0
\end{align}
So
$$a-b=\dfrac{\pi}{2}.$$
Now, consider the constraint $a<b\iff a-b<0$. Since $a-b=\dfrac{\pi}{2}>0$, no $a$ and $b$ satisfying
$$a-b=\dfrac{\pi}{2}.$$
So, I can't determine $\frac{b}{a}$.
It's my attempt correct? I confused to find $\frac{b}{a}$.
 A: This problem does not require any Math.
Instead, all that is required is visualization and meta-cheating.
The $\cos(a) + \cos(b) = 0$ constraint is equivalent to saying that the angles $a$ and $b$ are reflections of each other, about the $y$ axis.
A visualization of the sine and cosine functions against the backdrop of the unit circle will bear this out.  If you wish, you can consider that
$\cos(\pi - x) = \cos(\pi)\cos(x) - \sin(\pi)\sin(x) = -\cos(x)$.
So, you have that because of this reflection visualization, you must have that $\sin(a) = \sin(b).$
Edit
Technically, you must have that $\sin(a) = \pm \sin(b)$, but the distinction won't be relevant.
From there, it is time to meta-cheat, which means making the assumption that the problem is solvable (else why would the problem be posed).
Clearly, one possible solution is $a = \pi/4, b = 3\pi/4$, which leads to the solution $(b/a) = 3.$

Edit
More formally, it is clear that since the sine function is bounded above by $(+1)$, a consequence of the equation $\sin(a) + \sin(b) = \sqrt{2}$ is that you must have that $\sin(a)$ and $\sin(b)$ are both positive.
This makes it clear that you must have $a = \pi/4, b = 3\pi/4$.
