Cos (a+b) sin a=-3/5 sin b=-5/13 a & b in Q3

Sin(a) and sin(b) are both negative which means that a and b have to be in the 3rd or 4th quadrant.

  $$cos(a+b)\\\\ =cos(a)cos(b)-sin(a)sin(b)\\\\ =\pm\frac{4}{5}\times \pm \frac{12}{13}- \;\;\frac{-3}{5}\times \frac{-5}{13}\\\\ =\pm\frac{48}{65}- \;\;\frac{15}{65}\\\\ =\frac{-15-48}{65}\;\;or\;\;\frac{-15+48}{65}\\\\ =\frac{-63}{65}\;\;or\;\;\frac{33}{65}\\\\$$

Cos (a+b) sin a=-3/5 sin b=-5/13 a & b in Q3  ?

$$\boxed{\cos(a+b)=\;?} \quad \sin{(a)} =-{3\over5} \qquad \sin {(b)}=-{5\over13}$$

$$\cos{(a+b)}=\cos{(a)}*\cos(b)-sin{(a)}*\sin{(b)}$$

cos(a)=?   and   cos(b)=?

$$\textstyle{ \cos{(a)}=\sqrt{1-\sin^2{(a)}}= \sqrt{1-({3\over5})^2}= {\sqrt{5^2-3^2}\over5}={\sqrt{16}\over5}={\pm4\over5}=\pm{4\over5} }$$

$$\textstyle{ \cos{(b)}=\sqrt{1-\sin^2{(b)}}= \sqrt{1-({5\over13})^2}= {\sqrt{13^2-5^2}\over13}={\sqrt{144}\over13}={\pm12\over13}=\pm{12\over13} }$$

$$\cos{(a+b)}=\pm ({4\over5}) \times ({12\over13}) - (-{3\over5}) \times (-{5\over13})$$

$$\cos{(a+b)}=\pm ({4\over5}) \times ({12\over13}) - ({3\over5}) \times ({5\over13})$$

$$\cos{(a+b)}=({\pm(4*12)\over5*13}) - ({3*5\over5*13})$$

$$\cos{(a+b)}={\pm(4*12)-3*5\over5*13}$$

$$\cos{(a+b)}={\pm48-15\over65}$$

$$\text{1.) }\cos{(a+b)}={48-15\over65}={33\over65}=0.50769230769$$

$$\text{2.) }\cos{(a+b)}={-48-15\over65}=-{63\over65}=-0.96923076923$$