若sinθ=α θ∈[0,π/2] α=1/2 + 1/6 + 1/12 + 1/20 + 1/30

a=1/(1*2)+1/(2*3)+…+1/(12*13)
=1-1/2+1/2-1/3+1/3-1/4+…+1/11-1/12+1/12-1/13
=1-1/13
=12/13
sinθ=a=12/13
2*sinθ/2*cosθ/2=sinθ=a=12/13
sinθ/2*cosθ/2=6/13 (sinθ/2)^2=4/13
tanθ/2=2/3

等于2/3

sinθ=α=12/13
2*sinθ/2*cosθ/2=sinθ=a=12/13
sinθ/2*cosθ/2=6/13 (sinθ/2)^2=4/13
==>tanθ/2=2/3

解答：
a=1/(1*2)+1/(2*3)+…+1/(12*13)
=(1-1/2)+(1/2-1/3)+…(+1/12-1/13 )
=1-1/13
=12/13.
sinθ=a=12/13 ,
∴cosθ=√[1-(12/13)^2]=5/12.
∴tan(θ/2)=(1-cosθ)/sinθ=(1-5/13)/(12/13)=2/3,
或：tan(θ/2)=sinθ/(1+cosθ)=(12/13)/(1+5/13)=2/3.

α=1/2 + 1/6 + 1/12 + 1/20 + 1/30 +1/42 +1/56 + 1/72 + 1/90 + 1/110 + 1/132 + 1/156=0.923=12/13
sinθ=2tanθ/2/(1+(tanθ/2)^2)=12/13
==>12(tanθ/2)^2-26tanθ/2+12=0
==>tanθ/2=2/3

1/2=1/(1*2)=1/1-1/2
1/6=1/(2*3)=1/2-1/3
1/121/(3*4)=1/3-1/4
……
1/132=1/(11*12)=1/11-1/12
1/156=1/(12*13)=1*12-1/13
sinθ=α=1/2 + 1/6 + 1/12