为什么cos(α+5π/12)=sin(π/12-α)?为什么cos(π/3-α)=sin(α+π/6)?
为什么cos(α+5π/12)=sin(π/12-α)?为什么cos(π/3-α)=sin(α+π/6)?
sin²α+√3sinαcosα-2cos²α=0,α∈(π/6,5π/12),求:(1)sin(2
已知(3sinα+cosα)/(3cosα-sinα)=2,则2-3sin(α-3π)sin(1.5π-α)-[cos(
已知α=7/12π,那么cosα√(1-sinα)/(1+sinα)+sinα√(1-cosα)/(1-cosα)=
已知(sinα-cosα)/(sinα+cosα)=1/3,则cos^4(π/3+α)-cos^4(π/6-α)的值为
若(sinα+cosα)/(sinα-cosα)=2 则sin(α-5π)·sin(3π/2-α)=?
若sinα+cosαsinα−cosα=2,则sin(α-5π)•sin(3π2-α)等于( )
化简:sin(π/4-α)cos(π/3-α)-cos(π/6+α)sin(π/4+α)=
1.化简sin²α+cosαcos(π/3+α)-sin²(π/6-α)
已知cos(π/2+α)=sin(α-π/2) 求sin(π-α)-cos(π+α)/cos(5π/2 -α)+2sin
已知 sinα+2cos(5π/2+α)/cos(π-α)-sin(π/2-α)=-1/4 求(sinα+cosα)平方
已知α∈(π,2π) sinα+cosα=1/5 求sinα*cosα sinα-cosα