Trigonometry and Reflex Angles? I get $16/65$ for this question, when the answers say $56/65$. I'm guessing it's because of the reflex angles part: If $A$ and $B$ are both reflex angles, and given $\cos A=3/5$ and $\tan B=12/5$, find the exact value of $\sin(A-B)$. Any help would be appreciated, thanks.

Note that we are working with standard $3,4,5$ and $5,12,13$ triangles. It is therefore a question of taking care over the signs of all the trigonometric

A reflex angle $A$ has $\sin A \le 0$ - if in doubt sketch the curve.

So for the angles given we must have $\sin A = -\frac 45$ and since $\tan B =\frac {\sin B}{\cos B}$ is positive, we must have $\cos B \lt 0$ so that $\sin B=-\frac {12}{13}$ and $\cos B=-\frac 5{13}$.

These can be plugged into the standard formula $\sin (A-B)=\sin A\cos B-\cos A\sin B$, again taking care over signs.