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

$sin(a+b)$

$=sinacosb+cosasinb$

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

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

$cos(a-b)$

$=cosacosb+sinasinb$

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

$=(2sqrt15+sqrt5)/12$