Question

设锐角三角形ABC的内角A,B,C的对应边分别为a,b,c,a=2bsinA

(1)求B的大小 （2）求cosA+sinC的取值范围

Answer 1

解:∵a=2bsinA,∴a/sinA=2b
又sinB=b/(a/sinA)=b/2b=1/2,∴B=30°.

cosA+sinC=cos[180°-(B+C)]+sinC=cos(150°-c)+sinc
=cos[90°+(60°-c)]+sinC
=-sin(60°-c)+sinC
=-(sin60°cosc-cos60°sinc)+sinc
=-(√3/2)cosc+(3/2)sinC
=(3/2)[sinC-(√3/3)cosC]
=(3/2)[sinC-tan30°cosc]
=(3/2)(1/cos30°)(sinccos30°-coscsin30°)
=(√3)sin(c-30°)
∵△ABC是锐角三角形,∴0°<C<90°,故-30°<C-30°<60°
∴-1/2<sin(C-30°)<√3/2
∴-√3/2<cosA+sinC<3/2.

Answer 2

不好意思我手机不能打符号