Question

sin（a-b）=3/5,sin（a+b）=-3/5,a-b属于（π／2,π）a＋b属于（3π／2,2π）,则cos2b

Answer 1

因为sin(a+b)=-3/5,sin(a-b)=3/5,且a-b∈（π/2,π),a+b∈（3π/2,2π),

则cos(a+b)=4/5,cos(a-b)=-4/5,

所以cos2b=cos[(a+b)-(a-b)]

=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)

=4/5×（-4/5）+（-3/5）×（3/5）

=-16/25-9/25

=-1.