Question

Question #c4d85

Answer 1

Given that both `a and b in [pi/2,pi]` means they are in quadrant `II`

So `sin a and sinb -> +ve`

but `cosa and cosb -> -ve`

Given `sina=2/3`

So
`cosa=-sqrt(1-(2/3)^2)=-sqrt5/3`

Given `cosb=-1/4`

So
`sinb=sqrt(1-(-1/4)^2)=sqrt15/4`

Hence

`sin(a+b)`

`=sinacosb+cosasinb`

`=2/3*(-1/4)+(-sqrt5/3)*sqrt15/4`

`=-(2+5sqrt3)/12`

`cos(a-b)`

`=cosacosb+sinasinb`

`=(-sqrt5/3)*(-1/4)+2/3*sqrt15/4`

`=(2sqrt15+sqrt5)/12`