在△ABC中,若sinB*sinC=COS^2(A/2),判断并证明△ABC的形状 感激!
在△ABC中,若sinB*sinC=COS^2(A/2),判断并证明△ABC的形状 感激!
数学人气:424 ℃时间:2019-10-19 22:38:16
优质解答
cos^2(A/2)=(1+cosA)/2,因此cosA=2sinB×sinC -1,又因A+B+C=180,故cos(180-B-C)=2sinB×sinC -1,将cos(180-B-C)展开,得cos(180-B-C)=cos180×cos(B+C)+sin180×sin(B+C)=-cos(B+C)+0=-cosB×cosC+sinB×sinC带入原式...
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