已知cosa=-9/41,且π<a<3π/2,求tan(π/4-a)的值(详解)

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已知cosa=-9/41,且π<a<3π/2,求tan(π/4-a)的值(详解)
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已知cosa=-9/41,且π<a<3π/2,求tan(π/4-a)的值(详解)
已知cosa=-9/41,且π<a<3π/2,求tan(π/4-a)的值(详解)

已知cosa=-9/41,且π<a<3π/2,求tan(π/4-a)的值(详解)
cosa=-9/41,
sin²a=1-cos²a=1-(9/41)²=(41²-9²)/41²=(41+9)(41-9)/41²=(50*32)/41²
π<a<3π/2
sina<0
sina=-40/41
tana=-40/9
tan(π/4-a)
=[tan(π/4)-tana]/[1+tan(π/4)tana]
=(1-tana)/(1+tana)
=(49/9)/(-31/9)
=-49/31

cosa=-9/41 ,sina=-40/41 , tana=40/9
tan(π/4-a)=(1-tana)/(1+tana)=(-31/9)/(49/9)=-31/49