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