Question

If `sin A = -5/13, pi < A < (3pi)/2`, and `cos B = 24/25, (3pi)/2 < B < 2pi`, what is `sin(A+B)`?

Answer 1

`sin(A+B)=-36/325`

Explanation:

As `sinA` is negative and is in `Q3`, `cosA` is negative and `cosA=-sqrt(1-(-5/13)^2)=-sqrt(1-25/169)=-sqrt(144/169)=-12/13`

Further, `cosB` is positive and is in `Q4`, `sinB` is negative and `sinB=-sqrt(1-(24/25)^2)=-sqrt(1-576/625)=-sqrt(49/625)=-7/25`

Hence `sin(A+B)=sinAcosB×cosAsinB`

= `-5/13×24/25+(-12/13)×(-7/25)`

= `-120/325+84/325=-36/325`