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I was solving this question which said $$ \sin x \cos y = -1/2$$ $$\cos x \sin y = 1/2$$ and we had to solve for $x$ and $y$.

One thing that I deduced was that if I simply add the two equation then I would get $$\sin x\cos y + \cos x\sin y=0$$ $$\Rightarrow \sin(x+y)=0$$ $$\Rightarrow x+y= n\pi$$

Now I can substitute either $x$ or $y$ back in the equation and can easily find the answer.

This is one single way of solving the equation. Can there be some more intuitive ways of reaching the answer? This is not a kind of doubt, but I just want to enhance my horizons regarding different approaches to a single problem.

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    You could divide them to get $\tan x = -\tan y$. – 2017-01-03
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    You have also to solve : $\cos x \sin(n\pi -x)=1/2$ – 2017-01-03

3 Answers 3

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Hint

Multiply both:

$$(2\sin x \cos x)( 2\sin y \cos y) =-1 \Rightarrow \sin(2x)\sin (2y)=-1$$

and use that $-1 \le \sin \alpha \le 1$

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    Sneaky trick with the bounds on $sin$, I love this answer! – 2017-01-03
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By dividing both sides of the first equation by the second, we obtain

$$\tan(x)=-\tan(y)$$

from which we find that $x=\ell \pi-y$

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Alternatively, continuing from where you left off, you have already $$x+y=n\pi$$

But also you have $$\sin x\cos y-\cos x\sin y=-1\implies\sin(x-y)=-1$$

$$\implies x-y=-\frac{\pi}{2}+m\cdot2\pi$$

Now solve simultaneously and you have $$x=-\frac{\pi}{4}+\frac{2m+n}{2}\pi$$

And $$y=\frac{\pi}{4}+\frac{n-2m}{2}\pi$$

Now choose pairs of $m,n\in \mathbb{Z}$