a·b=(1,cos(2α))·(2,1)=2+cos(2α)c·d=(4sinα,1)·(sinα/2,1)=2sinα^2+11a·b-c·d=2+cos(2α)-2sinα^2-1=1+cos(2α)-2sinα^2=cos(2α)+cos(2α)=2cos(2α)α∈(0,π/4),即:2α∈(0,π/2)故:2cos(2α)∈(0,2),即:a·b-c·d∈(0,2)2f(a·b)=|2+cos(2α)-1|=2|cosα^2|=2cosα^2f(c·d)=|2sinα^2+1-1|=2|sinα^2|=2sinα^2α∈(0,π/4),0sinα^2即:f...
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