化简:[sin(π+3α)cos(-α+3π)sin(α-3π/2)cos(11π/2-α)]÷[cos(3α+9π/2

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.
 
 
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