化简：（1）sin[α+(2n+1)π]+sin[α−(2n+1)π]sin(α+2nπ)•cos(α−2nπ)（2）1−cos4α−sin4α1−cos6α−sin6α．

Question

化简：
（1）(sin[α+(2n+1)π]+sin[α−(2n+1)π])/(sin(α+2nπ)•cos(α−2nπ))
（2）(1−cos4α−sin4α)/(1−cos6α−sin6α)．

Answer 1

原式=(sin(2nπ+π+α)+sin(−2nπ−π+α))/(sin(2nπ+α)•cos(−2nπ+α))=(sin(π+α)+sin(−π+α))/(sinα•cosα)
=(−sinα−sin(π−α))/(sinα•cosα)=(−2sinα)/(sinα•cosα)=−(2)/(cosα)
2）原式=((cos2α+sin2α)2−cos4α−sin4α)/((cos2α+sin2α)3−cos6α−sin6α)=(2cos2α•sin2α)/(3cos2αsin2α(cos2α+sin2α))＝(2)/(3)．