Question

已知sina=-4/5,a∈(π,3π/2),cosb=-12/13,b∈(π/2,π)

求1.sin（a+b） 2.cos2a

Answer 1

sina=-4/5,a∈(π,3π/2)
cosa=-3/5
cosb=-12/13,b∈(π/2,π)
sinb=5/13

sin(a+b)
=sinacosb+cosasinb
=-4/5*(-12/13)+(-3/5)*5/13
=48/65-15/65
=33/65

cos2a
=1-2sin²a
=1-2*(-4/5)²
=1-32/25
=-725

Answer 2

1.∵sina=-4/5 a∈(π,3π/2)
∴cosa=-3/5
∵cosb=-12/13,b∈(π/2,π)
∴sinb=5/13
∴sin(a+b)=sinacosb+sinbcosa=48/65-15/65=33/65
2.∵cos2a=cos^(2)a-sin^(2)a=9/25-16/25=-7/25