怎样用matlab或Mathematica化简根式，用最简洁的表达式表示的那种！

Question

本人在求sin1度的表达式时得出一个等式，经过几天的化简后得到一个长串的式子，但苦于化简太繁杂而容易出错，有没有谁能化简的啊，感激不尽，sin1度=(3^0.5*i)/(8*((4*5^0.5-10*3^0.5-2*15^0.5+20)^0.5/64 -(6^0.5+2^0.5)*(5^0.5-1)/128 +(-8-3^0.5-15^0.5-(10-20^0.5)^0.5)^0.5/32)^(1/3)) - ((4*5^0.5-10*3^0.5-2*15^0.5+20)^0.5/64-(6^0.5+2^0.5)*(5^0.5-1)/128 +(-8-3^0.5-15^0.5-(10-20^0.5)^0.5)^0.5/32)^(1/3)/2 - 1/(8*((4*5^0.5-10*3^0.5-2*15^0.5+20)^0.5/64-(6^0.5+2^0.5)*(5^0.5-1)/128 +(-8-3^0.5-15^0.5-(10-20^0.5)^0.5)^0.5/32)^(1/3)) - (3^0.5*((4*5^0.5-10*3^0.5-2*15^0.5+20)^0.5/64-(6^0.5+2^0.5)*(5^0.5-1)/128 +(-8-3^0.5-15^0.5-(10-20^0.5)^0.5)^0.5/32)^(1/3)*i)/2。
专业表达形式为：
(1/4)*2^(1/2)*(-((2*(7 + 5^(1/2) + (30 + 6*5^(1/2))^(1/2))^(1/2) + (2* I)*(9 - 5^(1/2) - (30 + 6*5^(1/2))^(1/2))^(1/2))^(1/3) - 2)^2/(2*(7 + 5^(1/2) + (30 + 6*5^(1/2))^(1/2))^(1/2) + (2* I)*(9 - 5^(1/2) - (30 + 6*5^(1/2))^(1/2))^(1/2))^(1/3))^(1/ 2)

Answer 1

matlab里算的：
(3^0.5*i)/(8*((4*5^0.5-10*3^0.5-2*15^0.5+20)^0.5/64 -(6^0.5+2^0.5)*(5^0.5-1)/128 +(-8-3^0.5-15^0.5-(10-20^0.5)^0.5)^0.5/32)^(1/3)) - ((4*5^0.5-10*3^0.5-2*15^0.5+20)^0.5/64-(6^0.5+2^0.5)*(5^0.5-1)/128 +(-8-3^0.5-15^0.5-(10-20^0.5)^0.5)^0.5/32)^(1/3)/2 - 1/(8*((4*5^0.5-10*3^0.5-2*15^0.5+20)^0.5/64-(6^0.5+2^0.5)*(5^0.5-1)/128 +(-8-3^0.5-15^0.5-(10-20^0.5)^0.5)^0.5/32)^(1/3)) - (3^0.5*((4*5^0.5-10*3^0.5-2*15^0.5+20)^0.5/64-(6^0.5+2^0.5)*(5^0.5-1)/128 +(-8-3^0.5-15^0.5-(10-20^0.5)^0.5)^0.5/32)^(1/3)*i)/2

ans =

0.0175 - 0.0000i
就是sin(1°) =0.0175啦，望采纳~追问

你这个等式跟我的一样啊，你自己检查一遍，帮忙想个办法把虚数单位i去掉啊！

追答

本来式子里不就是有虚数单位么，而且最后不是算出虚部为零么？

追问

这个我知道，但是化简到最后不应该是指剩下实数部分吗？那,实数部分的表达式时怎么样的呀?