Question

sinA+sinB+sinC>=cosA+cosB+cosC

Answer 1

证:

∵△ABC为锐角三角形,∴A+B>90°

得A>90°-B

∴sinA>sin(90°-B)=cosB,即

sinA>cosB,同理可得

sinB>cosC,

sinC>cosA

上面三式相加:sinA+sinB+sinC>cosA+cosB+cosC

所以在锐角三角形ABC中,求证sinA+sinB+sinC＞cosA+cosB+cosC,得证