化简:[sin(π+3α)cos(-α+3π)sin(α-3π/2)cos(11π/2-α)]÷[cos(3α+9π/2
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/18 11:56:45
化简:[sin(π+3α)cos(-α+3π)sin(α-3π/2)cos(11π/2-α)]÷[cos(3α+9π/2)sin(-α-π)
[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.
=[-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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