第九题求详细过程
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∵(tanπ/4+tanθ/2)/(1-tanπ/4tanθ/2)=2
(1+tanθ/2)/(1-tanθ/2)=2
2-2tanθ/2=1+tanθ/2
tanθ/2=1/3
∴tanθ=(2tanθ/2)/(1-tan²θ/2)
=(2/3)/(1-1/9)
=2/3 ×9/8
=3/4
∵θ是锐角
∴sinθ=3/5
cosθ=4/5
∴cos(π/3+θ)
=cosπ/3cosθ-sinπ/3sinθ
=1/2cosθ-√3/2sinθ)
=1/2×4/5-√3/2×3/5)
=4/10-3√3/10
=(4-3√3)/10
(1+tanθ/2)/(1-tanθ/2)=2
2-2tanθ/2=1+tanθ/2
tanθ/2=1/3
∴tanθ=(2tanθ/2)/(1-tan²θ/2)
=(2/3)/(1-1/9)
=2/3 ×9/8
=3/4
∵θ是锐角
∴sinθ=3/5
cosθ=4/5
∴cos(π/3+θ)
=cosπ/3cosθ-sinπ/3sinθ
=1/2cosθ-√3/2sinθ)
=1/2×4/5-√3/2×3/5)
=4/10-3√3/10
=(4-3√3)/10
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tan[(π/2)+θ]=tan2[(π/4)+(θ/2)]
=-4/3
tan[(π/2)+θ]=-cotθ
cotθ=cosθ/sinθ=4/3
sin²θ+cos²θ=1
=-4/3
tan[(π/2)+θ]=-cotθ
cotθ=cosθ/sinθ=4/3
sin²θ+cos²θ=1
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