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△ABC中,cosA=1/3,求sin∧2(B+C)/2+cos∧2(B+C)的值
人气:316 ℃ 时间:2019-08-20 17:08:36
解答
[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
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