Question

设锐角三角形ABC的内角ABC的对边分别是abc;; a=2bsinA 《1》求B的大小 《2》求CosA+SINc的取值范围

如题设锐角三角形ABC的内角ABC的对边分别是abc;; a=2bsinA 《1》求B的大小 《2》求CosA+SINc的取值范围

Answer 1

a=2bsinA
a/sinA=2b=b/(1/2)
sinB=1/2
B=30度,或B=120度.
sinθ+sinφ = 2 sin[(θ+φ)/2] cos[(θ-φ)/2]
CosA+sinc
=2sin[(A+c)/2] cos[(A-c)/2] =2sin[(180-B)/2] cos[(A-c)/2]
=2sin(B/2)cos[(A-c)/2]
=2sin(60)cosK (k=A-c)/2)
A<180-60=120 0< C<180-60=120
0<(A-C)/2<60
1/2<cos(A-C)/2<1
CosA+sinc=2sin(60)cosK =根号3cosK
(根号3)/2<CosA+sinc<根号3

或CosA+sinc=2sin(B/2)cos[(A-c)/2]
CosA+sinc=2sin(30/2)cosK
=2根号[(1-cos30)/2]cosk
=根号[2-根号3]cosk

A<180-30=150 0< C<180-30=150
0<(A-C)/2<75
cos75<cos(A-C)/2<1
cos75=cos(30+45)=cos30cos45-sin30sin45=[(根号6)-(根号2)]/4

CosA+sinc=2根号[(1-cos30)/2]cosk

[(根号6)-(根号2)]*根号[2-根号3]/4<CosA+sinc<2根号[(1-cos30)/2]
[根号(12-6根号3)-根号(4-2根号3)]/4<CosA+sinc<根号[2-根号3]