Question

已知在三角形ABC中,B=60,且1/cosA+1/cosC=-2根号2,求cos（A-C)的值

能不能不用积化和差，和差化积

Answer 1

利用积化和差和和差化积
解：B=60,A+C=120
1/cosA+1/cosC=-2√2
(cosA+cosC)/(cosAcosC)=-2√2
cosA+cosC=2cos[(A+C)/2]cos[(A-C)/2]=cos[(A-C)/2]
cosAcosC=-1/2[cos(A+C)+cos(A-C)]=-1/2[-1/2+cos(A-C)]
令cos[(A-C)/2]=t
t/(-1/2)[-1/2+2t²-1]=-2√2
化简4t²-√2t-3=0
公式法解出t=-√2/2或3√2/4>1（舍去）
cos[(A-C)/2]=-√2/2
cos(A-C)=2cos²[(A-C)/2]-1=2×1/2-1=0
参考

追问

能不能不用积化和差，和差化积

Answer 2

已知在三角形ABC中,B=60°
则A+C=180°-B=120°
又知1/cosA+1/cosC=-2根号2
即cosA+cosC=-2√2cosAcosC
2cos[(A+C)/2)]cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
2cos60°cos[(A-C)/2]=-√2[cos120°+cos(A-C)]
cos[(A-C)/2]=-√2[-1/2+2cos²(A-C)/2-1]
2√2cos²[(A-C)/2]+cos[(A-C)/2]-3√2/2=0
4cos²[(A-C)/2]+√2cos[(A-C)/2]-3=0
解得cos[(A-C)/2]=-3√2/4<-1(舍去) 或cos[(A-C)/2]=√2/2
所以cos(A-C)=2cos²(A-C)/2-1=0

追问

能不能不用积化和差，和差化积

Answer 3

∵ B=60°, A+C=120°
1/cosA+1/cosC= - 2√2
∴ cosA+cosC=-2√2cosAcosC （左边用和差化积，右边用积化和差）
2cos[(A+C)/2]cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
cos[(A-C)/2]=-√2[-1/2+cos(A-C)]
=-√2{2[cos(A-C)/2]^2-3/2}
设：cos[(A-C)/2]=t
则：t=-√2{2t^2-3/2}
解得：t=√2/2
cos[(A-C)/2]=√2/2
cos(A-C)=2【cos[(A-C)/2】^2-1=2(√2/2)^2-1=1-1=0

追问

能不能不用积化和差，和差化积

Answer 4

B=60°, A+C=120°
1/cosA+1/cosC= - 2√2
∴ cosA+cosC=-2√2cosAcosC （左边用和差化积，右边用积化和差）
2cos[(A+C)/2]cos[(A-C)/2]=-√2[cos(A+C)+cos(A-C)]
cos[(A-C)/2]=-√2[-1/2+cos(A-C)]
=-√2{2[cos(A-C)/2]^2-3/2}
设：cos[(A-C)/2]=t
则：t=-√2{2t^2-3/2}
解得：t=√2/2
cos[(A-C)/2]=√2/2
cos(A-C)=2【cos[(A-C)/2】^2-1=2(√2/2)^2-1=1-1=0

Answer 5

1/cosA+1/cosC= -2√2
(cosC+cosA)/cosAcosC=-2√2
[ cos[(C+A)/2+(C-A)/2] +cos[(C+A)/2 -(C-A)/2] ] /cosAcosC=-2√2
2cos[(C+A)/2]cos[(C-A)/2]/cosAcosC =-2√2
cos[(C-A)/2]/cosAcosC=-2√2
cos[(C-A)/2]/cosAcosC=-2√2
cos[(C-A)/2]/[[cos(A+C)+cos(A-C)]/2 ]=-2√2
2cos[(C-A)/2]=[cos120+cos(A-C)](-2√2)
cos[(C-A)/2]=√2/2-√2cos(A-C)
设t=cos[(C-A)/2]
t=√2/2 - √2 *(2t^2-1)
√2t+4t^2=3
(2t+√2/4)^2=3+2/16
2t+√2/4=√51/4 或 2t+√2/4=-√51/4
t=(√51-√2)/8 t=(-√51-√2)/8
cos(A-C)=2t^2-1
cos(A-C)=(53-2√102)/64 -1 或 cos(A-C)=(53+2√102)/64 -1

追问

能不能不用积化和差，和差化积

追答

[ cos[(C+A)/2+(C-A)/2] +cos[(C+A)/2 -(C-A)/2] ] /cosAcosC=-2√2
和差化积2cos[(C+A)/2]cos[(C-A)/2]/cosAcosC=-2√2

cos[(C-A)/2]/cosAcosC=-2√2
积化和差cos[(C-A)/2]/[[cos(A+C)+cos(A-C)]/2 ]=-2√2