三角函数高一两小题已知sin x/2 - 2cos x/2 =0(1)求tanx的值(2)求 cos2x/√(2)cos

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  • (1)sin x/2 - 2cos x/2 =0 1式

    (sin x/2 )^2- (cos x/2)^2 =1 2式

    由1式和2式联立得

    sin x/2=2根5 cos x/2 =根5

    或sin x/2=-2根5 cos x/2 =-根5

    tan x/2=(sin x/2)/(cos x/2)=2

    tanx=(2tan x/2)/(1-(tan x/2)^2)=-4/3

    (2)cos2x/√(2)cos(π/4+x)*sinx

    =(2(cos x)^2-1)/(根2(cos(π/4)cos x-sin(π/4)sin x))*sinx

    =(2(cos x)^2-1)/(cos x-sin x)*sin x

    因为sin x/2=2根5 cos x/2 =根5

    或sin x/2=-2根5 cos x/2 =-根5

    所以(2(cos x)^2-1)/(cos x-sin x)*sin x

    =(2(根5)^2-1)/(根5-2根5)*2根5或(2(-根5)^2-1)/(-根5-(-2根5))*-2根5

    =-18