题目

△ABC中,C=45°,求sinA+sinB的最大值

提问时间：2021-03-20

答案

sinA+sinB
= sin((A+B)/2+(A-B)/2) + sin((A+B)/2-(A-B)/2)
= sin((A+B)/2)*cos((A-B)/2) + cos((A+B)/2)*sin((A-B)/2) + sin((A+B)/2)*cos((A-B)/2) - cos((A+B)/2)*sin((A-B)/2)
= 2*sin((A+B)/2)*cos((A-B)/2)
由于A+B = π-π/4 = 3π/4,
= 2 * sin(3π/8) * cos((A-B)/2)
显然,A = B的时候可以取最大值2*sin(3π/8)