在△ABC中,A、B、C对应边分别为a、b、c,已知2[sin((A+B)/2)]^2=1+cos2C.(1)求C的大小;(2)若c^2=2b^2-2a^2,求cos2A-cos2B的值。
(1)2[sin((A+B)/2)]^2=1+cos2C →cos(A+B)+cos2C=0 →2(cosC)^2-cosC-1=0 ∴cosC=-1/2或cosC=1/2(舍) 即C=2π/3. (2)c^2=2b^2-2a^2 →(sinC)^2=2(sinB)^2-2(sinA)^2 故cos2A-cos2B=3/4。
(1)2[sin((A+B)/2)]^2=1+cos2C,1-cos(A+B)=2(cos)^2,2(cosC)^2-cosC-1=0,cosC≠1,cosC=-1/2,C=120度
(2)c^2=2b^2-2a^2,由正弦定理,(sinC)^2=2(sinB-sinA)(sinB+sinA)
4(sin(A+B)/2)^2*(cos(A+B)/2)^2=8cos(B+A)/2*sin(B-A)/2*sin(A+B)/2*cos(B-A)/2
2sin(B+A)/2*cos(B+A)/2=4sin(B-A)/2*cos(B-A)/2
sin(B+A)=2sin(B-A)
sin(B-A)/2=(1/2)sin60°=(√3)/4
cos2A-cos2B=-2sin(A+B)sin(A-B)=-2*((√3)/2)*(-(√3)/4)=3/4。
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