在△ABC中,BC=根号5,AC=3,4cos2A－cos2C=3,1.求AB的值
1个回答
a=√5,b=3
由4cos2A-cos2C=3可得：
4-8sin^2A-1+2sin^C=3
2sinA=sinC
所以由正弦定理：
absinC=bcsinA求出c
a2sinA=csinA
c=2a=2√5
由余弦定理可以求出角度：
cosB=(a^2+c^2-b^2)/2ac=(5+20-9)/20=4/5
B=arccos(4/5)
cosA=2√5/5 A=arccos(2√5/5)
cosC=-√5/5 C=π-arccos(√5/5)
2、
sin^2A=1-cos^2A=1-20/25=1/5
sinA=√5/5
sin(2A-π/4)=sin2Acosπ/4-cos2Asinπ/4
=( √2/2)sin2A-(√2/2)cos2A
=√2sinAcosA-√2/2+√2sin^2A
=√2*2√5/5*√5/5-√2/2+√2/5
=2√2/5-√2/2+√2/5=3√2/5-√2/2
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