三角形中角B=45°求sinA+sinC的取值范围

1个回答

A+C=π-B=0.75π
sinA+sinC=sinA+sin(0.75π-A)=2sin[(A+0.75π-A)/2]cos[(A-0.75π+A)/2]
=2sin3π/8cos(3π/8-A）
当A=3π/8时,原式有最大值2sin（3π/8）=2[0.5(1-0.5^0.5)]^0.5
当A=0时,原式有最小值2sin（3π/8）cos（3π/8）=sin（3π/4）=0.5^0.5