2sin^2B/2+2sin^2C/2=1,试判断三角形ABC的形状 答案是等边三角形,
问题描述:
2sin^2B/2+2sin^2C/2=1,试判断三角形ABC的形状 答案是等边三角形,
答
∵2sin²(B/2)+2sin²(C/2)=1 ∴[2sin²(B/2)-1]+[2sin²(C/2)-1]=1-2 即 (-cosB)+(-cosC)=-1 ∴ cosB+cosC=1 ∴ 2cos(B/2+C/2)*cos(B/2-C/2)=1 即 2sin(A/2)*cos(B/...