已知tan(π/4+θ/2)=2,θ为锐角,则cos(π/3+θ)是

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已知tan(π/4+θ/2)=2,θ为锐角,则cos(π/3+θ)是已知tan(π/4+θ/2)=2,θ为锐角,则cos(π/3+θ)是已知tan(π/4+θ/2)=2,θ为锐角,则cos(π/3+θ

已知tan(π/4+θ/2)=2,θ为锐角,则cos(π/3+θ)是
已知tan(π/4+θ/2)=2,θ为锐角,则cos(π/3+θ)是

已知tan(π/4+θ/2)=2,θ为锐角,则cos(π/3+θ)是
令x=tan(θ/2),依题意得(1+x)/(1-x)=2,解之得x=1/3
sinθ=2x/(1+x^2)=3/5,而θ为锐角,所以cosθ=4/5
cos(π/3+θ)=cos(π/3)cosθ-sin(π/3)sinθ
=1/2*4/5-√3/2*3/5=(4-3√3)/10


tan(π/4+θ/2)
=(tanπ/4+tanθ/2)/(1-tanπ/4tanθ/2)
=(1+tanθ/2)/(1-tanθ/2)
=2
即(1+tanθ/2)/(1-tanθ/2)=2
解之得
tanθ/2=1/3
利用万能公式,可得
sinθ=2tanθ/2/(1+tan^2θ/2)=3/5
cosθ=(...

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tan(π/4+θ/2)
=(tanπ/4+tanθ/2)/(1-tanπ/4tanθ/2)
=(1+tanθ/2)/(1-tanθ/2)
=2
即(1+tanθ/2)/(1-tanθ/2)=2
解之得
tanθ/2=1/3
利用万能公式,可得
sinθ=2tanθ/2/(1+tan^2θ/2)=3/5
cosθ=(1-tan^2θ/2)/(1+tan^2θ/2)=4/5
∴cos(π/3+θ)
=cosπ/3cosθ-sinπ/3sinθ
=(1/2)*(4/5)-(√3/2)*( 3/5)
=2/5 +3√3/10

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