已知A、B、C是三角形ABC的三个内角,设y=2sinA/cosA+cos(b-c)
问题描述:
已知A、B、C是三角形ABC的三个内角,设y=2sinA/cosA+cos(b-c)
(1)证明:y=cotB+cotC
(2)若A=60度,求y的最小值
答
y=2sin(B+C)/(-cos(B+C)+cos(b-c))=2(SINBCOSC+COSBSINC)/2SINBSINC=COSB/SINB+COSC/SINC=cotB+cotCy=2sin60°/(cos60°+cos(b-c))cos(b-c)=cos(2B-120°)-120<2B-120<120°-1/2<cos(2B-120°)≤1cos(2B-120°...