题目

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

提问时间：2020-10-02

答案

[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