sin^2(α)+cosαcos(π/3+α)-sin^2(π/6-α)为定值,要证明
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sin^2(α)+cosαcos(π/3+α)-sin^2(π/6-α)为定值,要证明
sin^2(a)+cosa*cos(pi/3+a)-sin^2(pi/6-a)
=sin^2(a)+cosa*(cospi/3cosa-sinpi/3sina)
-[sinpi/6cosa-sinacospi/6]^2
=sin^2(a)+cosa(1/2cosa-根号3/2sina)
-[1/2cosa-根号3/2sina]^2
=sin^2(a)+[1/2cos^2(a)-根号3/2sinacosa]
-[1/4cos^2(a)-根号3/2sinacosa+3/4sin^2(a)]
=[sin^2(a)-3/4sin^2(a)]+[1/2cos^2(a)-1/4cos^2(a)]-[根号3/2sinacosa-根号3/2sinacosa]
=[1/4sin^2(a)]+[1/4cos^2(a)]
=1/4[sin^2(a)+cos^2(a)]
=1/4
即无论a为何值,sin^2(α)+cosαcos(π/3+α)-sin^2(π/6-α)恒为定值 1/4
=sin^2(a)+cosa*(cospi/3cosa-sinpi/3sina)
-[sinpi/6cosa-sinacospi/6]^2
=sin^2(a)+cosa(1/2cosa-根号3/2sina)
-[1/2cosa-根号3/2sina]^2
=sin^2(a)+[1/2cos^2(a)-根号3/2sinacosa]
-[1/4cos^2(a)-根号3/2sinacosa+3/4sin^2(a)]
=[sin^2(a)-3/4sin^2(a)]+[1/2cos^2(a)-1/4cos^2(a)]-[根号3/2sinacosa-根号3/2sinacosa]
=[1/4sin^2(a)]+[1/4cos^2(a)]
=1/4[sin^2(a)+cos^2(a)]
=1/4
即无论a为何值,sin^2(α)+cosαcos(π/3+α)-sin^2(π/6-α)恒为定值 1/4
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