Question

△ABC中,cosA=1/3,求sin∧2(B+C)/2+cos∧2(B+C)的值

Answer 1

[sin(B+C)/2]^2+[cos(B+C)]^2

[2倍角公式]

=[1-cos(B+C)]/2+[cos(B+C)]^2

cos(B+C)=cos(π-A)=-cosA=-1/3

=(1+1/3)/2+1/9

=7/9