locmsimon asked in 教育及參考書教學 · 9 years ago

史上最難數學題,自認高手請進(22分或以上)

圖片參考:http://imgcld.yimg.com/8/n/HA00145646/o/7011100700...

請解釋圖1步驟用左咩formula. 以及解釋圖2 step6點去step7

thanks

圖片參考:http://imgcld.yimg.com/8/n/HA00145646/o/7011100700...

Update:

可唔可以prove下你所指出既rules?

1 Answer

Rating
  • Fai
    Lv 4
    9 years ago
    Favorite Answer

    圖一:

    用左sin(A+B)=sinAcosB+sinBcosA

    &sin2A=2sinAcosA

    &cos2A=1-2(sinA)^2

    &(sinA)^2+(cosA)^2=1

    圖二:

    第五步到第六步:

    4d^2*(sinθ)^2/(sin2θ)^2 * (1-(sinθ)^2) remarks: (sinθ)^2+(cosθ)^2=1

    =4d^2(sinθ)^2/(sin2θ)^2 * (cosθ)^2

    第六步到第七步

    4d^2*(sinθ)^2*(cosθ)^2/(sin2θ)^2 remarks: 2sinθcosθ=sin2θ

    =4d^2*(sinθ)^2*(cosθ)^2/(2sinθcosθ)^2

    唔識再寄信比我啦!!!!

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