Suppose that $x$ and $y$ satisfy the equations

$$
\begin{aligned}
& 3 \sin x+4 \cos y=5 \\
& 4 \sin y+3 \cos x=2
\end{aligned}
$$

What is the value of $\sin (x+y)$ ?
We square the two given equations to obtain

$$
\begin{aligned}
& (3 \sin x+4 \cos y)^{2}=5^{2} \\
& (4 \sin y+3 \cos x)^{2}=2^{2}
\end{aligned}
$$

or

$$
\begin{aligned}
& 9 \sin ^{2} x+24 \sin x \cos y+16 \cos ^{2} y=25 \\
& 16 \sin ^{2} y+24 \sin y \cos x+9 \cos ^{2} x=4
\end{aligned}
$$

Adding these equations and re-arranging, we obtain

$$
9 \sin ^{2} x+9 \cos ^{2} x+16 \sin ^{2} y+16 \cos ^{2} y+24 \sin x \cos y+24 \cos x \sin y=29
$$

Since $\sin ^{2} \theta+\cos ^{2} \theta=1$ for every angle $\theta$, then

$$
9+16+24(\sin x \cos y+\cos x \sin y)=29
$$

from which we obtain

$$
\sin x \cos y+\cos x \sin y=\frac{4}{24}
$$

and so $\sin (x+y)=\frac{1}{6}$. (This uses the Useful Fact for Part A.)

(It is possible to solve for $\sin x, \cos x, \sin y$, and $\cos y$. Can you see an approach that would allow you to do this?)