有关正余弦定理的题目

1.在△ABC中，A+B+C=180°

y=cosB+cos[(180°-B)/2] ∴

= cosB+sin(B/2)

=1-2 sin? (B/2)+ sin(B/2)

=-2[sin(B/2)-1/4) ?+9/8

∵0°<B<180°

∴0°<B/2<90°

∴0< sin(B/2<1

-1/4<sin(B/2)-1/4<3/4

0< [sin(B/2)-1/4) ?<9/16

-9/8<-2 [sin(B/2)-1/4) ?<0

0<-2 [sin(B/2)-1/4) ?+9/8<9/8

∴y的取值范围是：（0，9/8）

5.(1)由正弦定理得：

(2sinA+sinC)cosB+sinBcosC=0

∴2sinAcosB+ sinCcosB+ sinBcosC

=2sinAcosB+sin(B+C)

=2sinAcosB+sinA=0

=>cosB=-1/2

∴B=120°

(2) ∵a+c=4

∴a=4-c,(a+c) ?=16

∴a?+c?=16-2ac

由余弦定理得：13＝a?+c?-2accosB= a?+c?+ac=16-ac

∴ac=a(4-a)=3

解得;a=1或a=3

正弦余弦定理题目求解，要过程，谢谢！！

解：

利用正弦定理

c/sinC=b/sinB=a/sinA=√3/(√3/2)=2

b=2sinB,c=2sinC=2sin(120°-B)

b+c

=2sinB+2sin(120°-B)

=2sinB+2sin120°cosB-2cos120°sinB

=3sinB+√3cosB

=2√3[sinB*(√3/2)+cosB*(1/2)]

=2√3(sinBcos30°+cosBsin30°)

=2√3sin(B+30°)

当B=60°时，b+c有最大值2√3

1、

2abcosC(2sinA-sinB)=2accosBsinB

bcosC(2sinA-sinB)=csinBcosB

sinBcosC(2sinA-sinB)=sinCsinBcosB

cosC(2sinA-sinB)=sinCcosB

2sinAcosC-sinBcosC=sinCcosB

2sinAcosC=sinBcosC+sinCcosB

2sinAcosC=sin(B+C)

2sinAcosC=sinA

2cosC=1

cosC=0.5

0<C<π

C=π/3

2、 设角ADC＝θ　则 设角BDC＝π-θ

AC^2=CD^2+AD^2-2CD*ADcosθ　=6-4*2^2^0.5cosθ

BC^2=CD^2+BD^2-2CD*BDcos(π-θ) =6-4*2^2^0.5cosθ(π-θ) =6+4*2^2^0.5cosθ

AC^2+BC^2=12

AB^2=AC^2+BC^2-2AC*BCcosπ/3

8=12-AC*BC

AC*BC=4

S=0.5*AC*BC*sin角ACB =0.5*4*cosπ/3=根号3