If tan A= 3/4 and cos B=-5/13, where A and B are both third-quadrant angles, find sin(A+B).

Question

Please explain how you found all the answers

Answer 1

make sketches of two right angles triangles in the third quadrants

since tanA = 3/4, the terminal arm of angle A ends at (-4,-3) , and r = 5
so sinA = -3/5, and cosA = -4/5
(you have a 3-4-5 right-angled triangle)

since cosB = -5/13 , the terminal arm of angle B ends at (-5,-12), and r = 13
so sinB = -12/13
(you have the 5-12-13 right-angled triangle)

we know
sin(A+B) = sinAcosB + cosAsinB
= (-3/5)(-5/13) + (-4/5)(-12/13)
= 15/65 + 48/65
= 63/65