Use the information given below to find sin(a+B)

Question

sin a = -3/5 in quadrant III

sin B = -4/5 in quadrant IV

Answer 1

sin a = -3/5 in QIII → cos a = -4/5

sin b = -45 in QIV → →cos b = -3/5

sin (a + b) = sin a cos b + cos a sin b Locate this formula and copy it!

Answer 2

sin(a+B)=sin(a)*cos(B)+cos(a)*sin(B)=-3/5*cos(B)-4/5*cos(a). This is due to substitution.

Next, consider each triangle is a 3-4-5 right triangle implying cos(a)=-4/5 and cos(B)=3/5 due to the follow:

1) Angle ‘a’ is in Quadrant lll,

2) Angle ‘B’ is in Quadrant lV.

Hence, sin(a+B)=-3/5*(3/5)+(-4/5)*(-4/5)=-9/25+16/25=7/25.