You are given that sin(A)=−9/41, with A in Quadrant III, and cos(B)=4/5, with B in Quadrant I. Find sin(A+B). Give your answer as a fraction.

Answer 1

sin(A+B) = -156/205

A involves a 9, 40 41 right triangle in 3rd quadrant 9^2 +40^2 = 41^2

B involves a 3,4, 5 right triangle in 1st quadrant 3^2 +4^2 = 5^2

sinA= -9/41 A=12.68+ 180 = 192.68 degrees cosA = -40/41

sinB = 3/5 B = 36.87 degrees cosB=4/5

A + B = 229.55 sin229.55 = -0.761

sin(A+B) = sinAcosB + sinBcosA = -9/41(4/5) + 3/5(-40/41) = -36/205 - 120/205 = -156/205

= about -0.761