在锐角三角形ABC中,证明1+cosA+cosB+cosC

1+cosA+cosB+cosC-(sinA+sinB+sinC)
=2[cos(A/2)]^2+2cos(B+C)/2*cos(B-C)/2-2[sin(A/2)*cos(A/2)+sin((B+C)/2)*cos((B-C)/2)]
=2[cos(A/2)*(cos(A/2)-sin(A/2))+cos((B-C)/2)*(cos((B+C)/2)-sin((B+C)/2))]
=2[cos(A/2)*(cos(A/2)-sin(A/2))+cos((B-C)/2)*(sin(A/2)-cos(A/2))]
=2(cos(A/2)-sin(A/2))*(cos(A/2)-cos((B-C)/2))
=-4(cos(A/2)-sin(A/2))*sin((A+B-C)/4)*sin((A+C-B)/4).①
由于A,B,C∈(0,π/2)
所以0

证明:∵△ABC为锐角三角形,∴A+B>90°
即A>90°-B,∴sinA>sin(90°-B)=cosB,即
sinA>cosB,同理
sinB>cosC,
sinC>cosA
上面三式相加:sinA+sinB+sinC>1+cosA+cosB+cosC