Question

if 270°<A<360°,90°<B<180°,cosA=5/13,tanB=(-15/8)then sin(A+B)=?​

Answer 1

Step-by-step explanation:

cosA=5/13

sinA=-√(1-cos^2A) (since theta is in 4th quadrant)

i.e. sinA=-√144/169

=-12/13

tanB=-15/8

sinB=+(15)/√{(-15)^2+(8)^2} (B is in Q2)

i.e. sinB=15/17

cosB=tanB/sinB

i.e. cosB=-15/8 * 17/15

cosB=17/8

now

sin(A+B)=sinAcosB+cosAsinB

sin(A+B)=(-12/13)(17/8)+(5/13)(15/17)

={1/13}(-51/2 + 75/17)

=(1/13)(-710/34)

=-710/442