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+...答:△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...
三角形ABC三个内角A,B,C所对边a,b,c,bcosA=√2a,则a/c取值范围答:也即sinB=√2tanA,由sinB的范围,0<tanA<=√2/2 而且sinB=√2tanA>tanA>sinA,因此B可以为钝角,实际上,B的范围是(0,π)又sinC=sin(A+B)=sinAcosB+sinBcosA=sinAcosB+√2sinA 因为B的取值范围是(0,π),cosB的范围是(-1,1),因此sinC/sinA的取值范围是(√2-1,√2+1)a/c=sinA/...