若tan(π/4-α)=3则tan2α等于
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若tan(π/4-α)=3则tan2α等于
解由tan(π/4-α)=3
得(tanπ/4-tanα)/(1+tanπ/4tanα)=3
即(1-tanα)/(1+tanα)=3
即1-tana=3+3tana
即4tana=-2
即tana=-1/2
即tan2α
=2tana/(1-tan^2a)
=2×(-1/2)/[1-(-1/2)^2]
=-1/(3/4)
=-4/3
再问: 那sin15°cos75°-cos15°sin105°该怎么算
再答: sin15°cos75°-cos15°sin105°
=sin15°cos75°-cos15°sin75°
=sin(15°-75°)
=-sin60°
=-√3/2.
得(tanπ/4-tanα)/(1+tanπ/4tanα)=3
即(1-tanα)/(1+tanα)=3
即1-tana=3+3tana
即4tana=-2
即tana=-1/2
即tan2α
=2tana/(1-tan^2a)
=2×(-1/2)/[1-(-1/2)^2]
=-1/(3/4)
=-4/3
再问: 那sin15°cos75°-cos15°sin105°该怎么算
再答: sin15°cos75°-cos15°sin105°
=sin15°cos75°-cos15°sin75°
=sin(15°-75°)
=-sin60°
=-√3/2.
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