Question #c4d85

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Answer 1 · 2017-03-07T07:28:24

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#