在△ABC中，内角A，B，C分别对应的边是a，b，c已知c=2，C=π/3。求sinA+sinB的...

Question

在△ABC中，内角A，B，C分别对应的边是a，b，c已知c=2，C=π/3。求sinA+sinB的取值范围

Answer 1

△ABC中,C=π/3,则0<=(A,B)<=2π/3
A=π-B-C=2π/3-B
sinA+sinB=sin(2π/3-B)+sinB=sin(2π/3)cosB-cos(2π/3)sinB+sinB=[(√3)cosB+3sinB]/2
=√3(cosB/2+√3sinB/2)=√3sin(B+π/6),
0<=(A,B)<=2π/3,则π/6<=B+π/6<=5π/6,1/2<=sin(B+π/6)<=1,
所以√3/2<=sinA+sinB<=√3

Answer 2

sinA+sinB
=sinA+sin(120-A)
=2sin[(A+120-A)/2]cos[(A-120+A)/2]
=2sin60cos(A-60)
= √3cos(A-60)
0 < A<120
-60<A<60
1/2 < cos(A-60)<1
即√3/2<√3cos(A-60)<√3
所以 √3/2<sinA+sinB)<√3

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Answer 3

sinA+sinB=sinA+sin(A+C)=sinA+sin(A+π/3)=sinA*3/2+cosA*√3/2=√3sin(A+π/6)
∵0<A<2π/3
∴π/6<A+π/6<5π/6
∴1/2<sin(A+π/6)≤1
∴√3/2<sinA+sinB≤√3

Answer 4

√3/2<sinA+sinB≤√3