有解析,望采纳
证明:
lg(sinBsinC)=lg((cos(A/2))^2)
sinBsinC=(cos(A/2))^2=(cosA+1)/2
2sinBsinC=-cos(B+C)+1
2sinBsinC=-cosBcosC+sinBsinC+1
cosBcosC+sinBsinC=1
cos(B-C)=1
B-C=0
B=C
所以△ABC是等腰三角形.
有解析,望采纳
证明:
lg(sinBsinC)=lg((cos(A/2))^2)
sinBsinC=(cos(A/2))^2=(cosA+1)/2
2sinBsinC=-cos(B+C)+1
2sinBsinC=-cosBcosC+sinBsinC+1
cosBcosC+sinBsinC=1
cos(B-C)=1
B-C=0
B=C
所以△ABC是等腰三角形.