史上最難數學題,自認高手請進(22分或以上)

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!!!!