$\cos(x)=-\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$. Calculate $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$. If $\cos(x)=  -\frac{24}{25}$ and $\tan(y) = \frac{9}{40}$ for $\frac{\pi}{2} < x < \pi$ and $\pi < y <\frac{3\pi}{2}$. What is the value of $\sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x) \cos(y) - \sin(x) \sin(y)$?

Solution:
If $\frac{\pi}{2} < x < \pi$ then right triangle with acute angle $x$ facing west at second quadrant, the hypotenuse must be positive and the only negative is the adjacent. 
$\cos(x)=  -\frac{24}{25}$ then $adjacent=-24$ and $front=7$ so $\sin(x) = \frac{7}{25}$.  
If $\pi < y <\frac{3\pi}{2}$ then right triangle with acute angle $x$ facing west at third quadrant, the front and adjacent are negative. If $\tan(y) = \frac{9}{40}$ then the front is $-9$, the adjacent is $-40$, and the hypotenuse is $41$. So $\sin(y) = -\frac{9}{41}$ and $\cos(y)=-\frac{40}{41}$
The right triangle that we consider is the one with the adjacent side on the $x$-axis right..?

You can observe that
$$
\lvert\sin x\rvert=\sqrt{1-\cos^2x}=\sqrt{1-\frac{24^2}{25^2}}=\frac{7}{25}
$$
Since $\pi/2<x<\pi$, we conclude that $\sin x=7/25$.
Also
$$
\lvert\cos y\rvert=\sqrt{\frac{1}{1+\tan^2y}}=\frac{40}{41}
$$
and from $\pi<y<3\pi/2$ we conclude that $\cos y=-40/41$. Thus
$$
\sin y=\tan y\cos y=-9/41
$$
