若x=π/12,则cos(2x-π)*cos[2(x-π)]=?

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若x=π/12,则cos(2x-π)*cos[2(x-π)]=?若x=π/12,则cos(2x-π)*cos[2(x-π)]=?若x=π/12,则cos(2x-π)*cos[2(x-π)]=?原式=c

若x=π/12,则cos(2x-π)*cos[2(x-π)]=?
若x=π/12,则cos(2x-π)*cos[2(x-π)]=?

若x=π/12,则cos(2x-π)*cos[2(x-π)]=?
原式=cos(-5π/6)*cos(-11π/6)
=cos(5π/6)*cos(11π/6)
=-cos²(5π/6)
=[cos(5π/3)-1]/2
=[-cos(2π/3)-1]/2
=(1/2-1)/2
=-1/4