若b²=ac,则cos(A-C)+cosB+cos2B=

如题所述

cos(A-C)+cosB+cos2B
=cos(A-C)-cos(A+B)+cos2B
=2sinA*sinC+cos2B,
而,b^2=ac,有
sin^2B=ainA*sinC,
cos2B=1-2*sin^2B,
所以,有
2sinA*sinC+cos2B
=2sinA*sinC+1-2sin^2B
=2*sin^B+1-2sin^2B
=1.
即,cos(A-C)+cosB+cos2B=1.
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