Question

sin(α+nπ)+sin(α-nπ)/sin(α+nπ)*cos(α-nπ)

要过程、谢谢！

Answer 1

若n是偶数
则nπ是2π的整数倍
所以原式=(sinx+sinx)/(sinx*cosx)=2sinx/(sinxcosx)=2/cosx
若n是奇数
则nπ=(2k+1)π是2kπ+π
原式=[sin(x+π)+sin(x-π)]/[sin(x+π)*cos(x-π)]
=[sin(x+π)-sin(π-x)]/[sin(x+π)*cos(π-x)]
=(-sinx-sinx)/[(-sinx*(-cosx)]
=-2sinx/(sinxcosx)
=-2/cosx
故有原式=2/cosx