Question

高中三角函数题

已知三角形ABC的周长为9，AC=3,4cos2A-cos2C=3
（1）求AB的值（2）求sin（A-π/4）的值
（求详解，要过程）

Answer 1

1、4cos2A-cos2C=3
=4[1-2(sinA)^2]-[1-2(sinC)^2]
3-8(sinA)^2+2(sinC)^2=3
(sinC)^2-4(sinA)^2=0,
sinC=2sinA,(三角形内正弦无负号，故舍去负值），
sinC/sinA=2,根据下弦定理，sinC/sinA=c/a,
c/a=2,c=2a,2a+a+3=9,a=2,c=4,
AB=c=4,
2、根据余弦定理，
a^2=c^2+b^2-2bccosA,
cosA=7/8,
sinA=√（1-49/64）=√15/8，
sin(A-π/4）=sinAcos(π/4)-cosAsin(π/4)
=(√30-7√2)/16.

Answer 2

1) 令AB=c, 则有：BC=9-c-3=6-c
4cos2A-cos2C=3
4(1-2sin^2 A)-(1-2sin^2 C)=3
4sin^2 A=sin^2 C
2sinA =sinC
又因：sinA/(6-c)=sinC/c
两式相除有：2(6-c)=c, 解得c=4
即AB=4
2）a=6-c=2
cosA=(3^2+4^2-2^2)/(2*3*4)=7/8
sinA=√(1-7^2/8^2)=√15/8
sin(A-π/4)=√2/2(sinA-cosA)=√2/16*(√15-7)