Trigonometry identities $\sin(A+B)$

Question

Given that $\cos A= 4/5$ and $\cos B= -12/13$ and that $A$ and $B$ are between $0$ and $180$ degrees, find the exact value of $\sin(A+B)$

With the given condition,

$\theta A$ is in the first quadrant, meaning $\sin A = 3/4 $

$\theta B$ is in the 2nd quadrant, meaning $\sin B= 5/13$

$$\sin (A+B) = \sin A \cos B + \cos A \sin B =\left(\frac{3}{4}\right)\left(\frac{-12}{13}\right) + \left(\frac{4}{5}\right)\left(\frac{5}{13}\right) = \frac{-5}{13}$$

Why am I wrong ?

Answer 1

As @user8734617 has said, the value of $\sin A$ was incorrectly calculated. From Pythagoras' Theorem, $$\sin A = \frac{\sqrt{5^2-4^2}}5=\frac35$$

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Also, from your answer, we can see that $$\sin(A+B)=-\sin B=\sin(-B)$$ This is true iff $$A+2B=k\pi\quad (k\in\mathbb{Z})$$ A quick check on your calculator shows that this is not the case.

Answer 2

With your hypothesis $$\sin A = \sqrt{1-\frac{4^2}{5^2}}= \sqrt{\frac{9}{5^2}}=3/5 $$