# If sinA=4/5 and cosB= -5/13, where A belongs to QI and B belongs to QIII, then find sin(A+B). How do you write the second half of the formula where cos becomes A and sin is B and how do the quadrants affect these?

Apr 30, 2018

A is in QI, so $\cos A = + \setminus \sqrt{1 - {\left(\frac{4}{5}\right)}^{2}} = \frac{3}{5}$ and B is in QIII, so $\sin B = - \sqrt{1 - {\left(- \frac{5}{13}\right)}^{2}} = - \frac{12}{13}$ so

$\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B = - \frac{56}{65}$

#### Explanation:

Are we being played by Socratic here? It's says this was asked 13 minutes ago in the feed but 3 years ago in the comment. Everybody should immediately recognize the Pythagorean Triples underlying this problem: ${3}^{2} + {4}^{2} = {5}^{2}$ and ${5}^{2} + {12}^{2} = {13}^{2.}$

${\cos}^{2} A + {\sin}^{2} A = 1$

$\cos A = \setminus \pm \setminus \sqrt{1 - {\sin}^{2} A}$

$\sin A = \frac{4}{5}$ and $A$ is in quadrant I, so the cosine is positive too.

$\cos A = \setminus \sqrt{1 - {\left(\frac{4}{5}\right)}^{2}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$

$\cos B = - \frac{5}{13}$ and $B$ is in quadrant III, so the sine is negative too.

$\sin B = - \setminus \sqrt{1 - {\left(- \frac{5}{13}\right)}^{2}} = - \frac{12}{13}$

$\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B$

$= \left(\frac{4}{5}\right) \left(- \frac{5}{13}\right) + \left(\frac{3}{5}\right) \left(- \frac{12}{13}\right) = - \frac{56}{65}$