Find the value of $\sin(B-A)$.

Question

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!

Answer 1

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 $\sin(B-A)$.

Answer 2

Note that, for the "before that" part: $\tan(2A)\equiv\frac{2\tan(A)}{1-\tan^2(A)}$. You know that $\tan(A)=\frac{15}{8},$ so sub. this into the formula.

Answer 3

Hint: Use pythagurus theorem and solve for $SinB, cosB,CosA, SinA$. Then substitute the values in your formula $Sin(B-A)$, you will get the answer.