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3/2sina...