I was trying to solve
$$ \sin (x-y)=3\sin x \cos y-1 \\ \sin (x+y)=-2\cos x \sin y$$
for $x,y$, so first I added both the equations and then subtracted both of them and solved these two equations the result I got was $$\frac{\tan y}{\tan x}=-\frac{1}{3}$$ but I am not able to further find the value of $x$ and $y$. Please can you help me solve this further?
Solve for $x,y$: $\sin (x-y)=3\sin x \cos y-1$ and $\sin (x+y)=-2\cos x \sin y$
trigonometry
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Best Answer
Let $a = x + y, b = x - y$. Then using the product-to-sum identities:
$$\sin b = 3 \sin \left( \frac{a+b}{2} \right) \cos \left( \frac{a-b}{2} \right) - 1 = \frac{3}{2} (\sin a + \sin b) - 1$$ $$\sin a = -2 \cos \left( \frac{a+b}{2} \right) \sin \left( \frac{a-b}{2} \right) = -(\sin a - \sin b)$$
and by a further substitution $p = \sin a, q = \sin b$:
$$2q = 3p + 3q - 2, p = q - p$$
which gives $0 = 3p + q - 2 = 3p + 2p - 2$, thus $p = \frac{2}{5}$ and $q = \frac{4}{5}$ and you can work your way backwards.