<s>[INST] Find the value of $\sin(B-A)$. If $A$ is an acute angle whose tangent is $\frac{15}{8}$ and $B$ is and obtuse angle whose sine is $\frac{12}{13}$, find $\sin (B-A)$.
[Without calculators]
I suppose I gotta use this formula: $\sin B \cos A - \cos B \sin A$
Before that they asked me to find $\tan 2A$ which is $\frac{-240}{161}$ and $\cos 2B$ which is $\frac{-119}{169}$
But I can't understand how to proceed....help please!
 [/INST] Label a right triangle with one acute angle $A$. Since $\tan A = \dfrac{15}{8}$ you can label the opposite side with length $15$ and the adjacent side with length $8$. Then the hypotenuse has length $17$ so that $\sin A = \dfrac{15}{17}$ and $\cos A = \dfrac{8}{17}$.
For $B$, use $\sin^2 B + \cos^2 B = 1$. Since $B$ is obtuse, its cosine is negative. Thus $\cos B = - \dfrac{5}{13}$. Now use the formula for $\sin(B-A)$.
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