$\sin(x) - \sin(y) = -\frac{1}{3}$, $\cos(x) - \cos(y) = \frac{1}{2}$, what is $\sin(x+y)$?

Question

If $\sin(x) - \sin(y) = -\frac{1}{3}$ and $\cos(x) - \cos(y) = \frac{1}{2}$, then what is $\sin(x+y)$?

---

Attempt:

$$ \sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y) $$

If we multiply the two "substraction identities" we get $$ (\sin(x) - \sin(y))(\cos(x) - \cos(y)) = -\frac{1}{6} $$ $$ \sin(x)\cos(x) + \sin(y) \cos(y) - \sin(x+y) = \frac{1}{6}$$ thus $$ \sin(x+y) =\sin(x)\cos(x) + \sin(y) \cos(y) - \frac{1}{6} $$

Next if I do a sum and organize and squaring we get:

$$ (\sin(x) + \cos(x))^{2} + (\sin(y) + \cos(y))^{2} - 2 (\sin(x) + \cos(x)) (\sin(y) + \cos(y)) = \frac{1}{36} $$ $$ 2 + 2 ( \sin(x) \cos(x) + \sin(y) \cos(y)) - 2 \left( \sin(x+y) + \cos(x-y) \right) = \frac{1}{36} $$

I have no idea after this.

---

Another method is I suspect that we have to solve for $\sin(x), \cos(x), \sin(y), \cos(y)$ instead of directly finding $\sin(x+y)$ algebraically.

Answer 1

If you start with the Sum-to-Product Identities, you can rewrite the given equations as $$2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)=-\frac{1}{3} \quad \text{and} \quad -2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)=\frac{1}{2}.$$

Dividing the second equation by the first one, we'll cancel out $\sin\left(\frac{x-y}{2}\right)$, and as a result we'll find the value of $\tan\left(\frac{x+y}{2}\right)$. Then to finish it up, use one of the half-angle substitutions: $$\sin\theta=\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}.$$

Answer 2

We have $$\cos \frac{x+y}{2} \sin \frac{x-y}{2}=-\frac{1}{6}$$ $$\sin \frac{x+y}{2} \sin \frac{x-y}{2}=-\frac{1}{4}$$ Thus $\tan \frac{x+y}{2}=\frac{3}{2}$. And hence $$\sin(x+y)=\frac{2\tan\frac{x+y}{2}}{1+\tan^2\frac{x+y}{2}}=\frac{12}{13}$$

Answer 3

HINT- $sinx-siny=2cos\frac{x+y}{2}sin\frac{x-y}{2}=-\frac{1}{3}$ ..........(1)

$cosx-cosy=-2sin\frac{x+y}{2}sin\frac{x-y}{2}=\frac{1}{2}$ ..........(2)

Divide to get $tan\frac{x+y}{2}=\frac{3}{2}$,Now use $sin\theta=2tan\frac{\theta}{2}/(1+tan^2\frac{\theta}{2})$