數學問題sin1*sin2.....sin88*sin89

sinx * sin(60-x) * sin(60+x) = 1/4 (sin3x) ...★
Sub x = 1 , 2 , 3 ...... , 29 , 30 to ★ :
(sin1)(sin59)(sin61) = (1/4)(sin3)
(sin2)(sin58)(sin62) = (1/4)(sin6)
...................................................
(sin29)(sin31)(sin89) = (1/4)(sin87)
(sin30)(sin30)(sin90) = (1/4)(sin90)
Product of L.H.S. = Product of R.H.S. :
(sin1)(sin2)...(sin30)^2 * (sin31)......(sin59)(sin61)......(sin89)(sin90)
= (1/4)^30 * (sin3)(sin6)(sin9)...(sin87)(sin90)
(sin1)(sin2)...(sin90) * (sin30)/(sin60)
= (1/4)^30 * (sin3)(sin6)(sin9)...(sin87)(sin90)
(sin1)(sin2)...(sin90)
= √3 * (1/4)^30 * (sin3)(sin6)(sin9)...(sin90)
Sub x = 3 , 6 , 9 ... , 27 , 30 to ★ :
Using the method above , we can get :
(sin3)(sin6)(sin9)...(sin30)^2 * (sin31)...(sin57)(sin63)...(sin87)(sin90)
=(1/4)^10 *(sin9)(sin18)(sin27)...(sin81)(sin90)
(sin3)(sin6)(sin9)...(sin87)(sin90)
= √3 * (1/4)^10 * (sin9)(sin18)(sin27)...(sin81)(sin90)
= √3 * (1/4)^10 * (sin9)(sin81)*(sin18)(sin72)*
(sin27)(sin63)*(sin36)(sin54)*(sin45)(sin90)
= √3 * (1/4)^10 * (sin9)(cos9)*(sin18)(cos18)*
(sin27)(cos27)*(sin36)(cos36) * (√2 /2) * (1)
= √3 * (1/4)^10 * (1/2)(sin18)*(1/2)(sin36)*
(1/2)(sin54)*(1/2)(sin72) * (√2 /2)
= √3 * (1/4)^10 * (sin18)(sin72)*(sin36)(sin54)*
(1/4)^2 * (√2 /2)
= √3 * (1/4)^10 * (sin18)(cos18)*(sin36)(cos36)*
(1/4)^2 * (√2 /2)
= √3 * (1/4)^10 * (1/2)(sin36)*(1/2)(sin72)*
(1/4)^2 * (√2 /2)
= √3 * (1/4)^10 * (1/4)^3 * (√2 /2) * (sin36)(sin72)
= (√6 /2) * (1/4)^13 * (sin36) * (sin72)
= (√6 /2) * (1/4)^13 * 2(sin18)(cos18)^2
Since sin36 = cos54
2(sin18)(cos18) = 4(cos18)^3 - 3cos18
2(sin18) = 4(cos18)^2 - 3
2(sin18) = 4[1-(sin18)^2] - 3
4(sin18)^2 + 2(sin18) - 1 = 0
sin18 = (√5 - 1) / 4
(cos18)^2 = 1 - (sin18)^2 = 1 - [(√5 - 1) / 4]^2
(cos18)^2 = (5 + √5) / 8
So (sin3)(sin6)(sin9)...(sin87)(sin90)

= (√6 /2) * (1/4)^13 * 2 (√5 -1)/4 * (5 + √5)/8
= (√6 /2) * (1/4)^15 * (√5 - 1)(√5 + 5)
= (1/4)^15 * 2√30
(sin1)(sin2)(sin3)...(sin89)(sin90)
= √3 * (1/4)^30 * (1/4)^15 * 2√30
= 6√10 / (4^45)