问题描述: 化简:[sin(π+3α)cos(-α+3π)sin(α-3π/2)cos(11π/2-α)]÷[cos(3α+9π/2)sin(-α-π) 1个回答 分类:数学 2014-11-23 问题解答: 我来补答 [sin(π+3 cos(-α+3π)sin(α-3π/2)cos(11π/2-α)]÷[cos(3α+9π/2)sin(-α-π)]=[-sin3α(-cosα) sin(α-3π/2+2π) cos(6π-(π/2+α))]÷[cos(3α+π/2)sin(-α-π+2π)]=[-sin3α(-cosα) sin(α+π/2) cos(-(π/2+α))]÷[-sin3αsin(-α+π)]=[-sin3α(-cosα) cosα(-sinα)]÷[-sin3αsinα]= cos²α. 展开全文阅读