Question

已知A+B=120°,求sinA+sinB的最大值.

Answer 1

A+B=120°，
sinA+sinB=sinA+sin(120°-A)= sinA+sin120°cos A -cos120°sin A
= sinA+√3/2 cos A+1/2 sin A=3/2 sin A+√3/2 cos A
=√3(√3/2 sin A+1/2 cos A)
=√3sin(A+30°)
因为0°<A<120°，30°<A+30°<150°
1/2<sin(A+30°)≤1，√3/2<√3sin(A+30°)≤√3.
所以sinA+sinB的范围是(√3/2, √3].
sinA+sinB的最大值是√3.

Answer 2

sinA+sinB
=sinA+sin(120-A)
=sinA+√3/2*cosA+1/2*sinA
=3/2*sinA+√3/2*cosA
=√3(√3/2*sinA+1/2*cosA)
=√3sin(A+30)

最大值为 √3.

Answer 3

sinA+sinB=sinA+sin(120-A)
=sinA+√3/2*cosA+1/2*sinA
=3/2*sinA+√3/2*cosA
=√3(√3/2*sinA+1/2*cosA)
=√3sin(A+π/6)
所以A=π/3时，原式最大，为√3

Answer 4

sinA+sinB=sin(120-B)+sinB
=sin120cosB-sinBcos120+sinB
=√3 /2 *cosB+3/2 *sinB
=√3 (√3 /2 *sinB+1/2 *cosB)
=√3 sin(B+30)
所以最大值是√3