Question

已知ab为锐角,且cosa=4/5,cos(a+b)=-1/3.求cosb

Answer 1

解

a,b是锐角

∴sina>0

∵cosa=4/5

由sin²a+cos²a=1

∴sina=√1-(4/5)²=3/5

∵a∈(0.π/2),b∈(0,π/2)

∴a+b∈(0.π)

∴sin(a+b)>0

∵cos(a+b)=-1/3

∴sin(a+b)=√1-(-1/3)²=2√2/3

∴cosb

=cos[(a+b)-a]

=cos(a+b)cosa+sin(a+b)sina

=-1/3×4/5+2√2/3×3/5

=(6√2-4)/15