化简三角函数[sin(α+pi)*cos(pi+α)*cos(α+2pi)]/[tan(pi+α)*cos^3(-α-pi)][sin(α+pi)*cos(pi+α)*cos(α+2pi)]/[tan(pi+α)*cos^3(-α-pi)]算出来最后等于1 ,
问题描述：
化简三角函数[sin(α+pi)*cos(pi+α)*cos(α+2pi)]/[tan(pi+α)*cos^3(-α-pi)]
[sin(α+pi)*cos(pi+α)*cos(α+2pi)]/[tan(pi+α)*cos^3(-α-pi)]
算出来最后等于1 ,
答
[sin(α+π)*cos(π+α)*cos(α+2π)]/[tan(π+α)*cos^3(-α-π)]
=[(-sinα)*(-cosα)*cosα]/[tanα*cos^3(α+π)]
=sinα*cosα*cosα]/[tanα*(-cosα)^3]
=sinα*cosα*cosα/[tanα*(-cosα)(-cosα)^2]
=sinα*cosα*cosα/[tanα*(-cosα)(cosα)^2]
=sinα/[tanα*(-cosα)]
=sinα/[-sinα]
=-1
答
原式=[(-sinα)(-cosα)cosα]/[tanα*(-cosα)^3]
=[sinα(cosα)^2]/[(sinα/cosα)(-cosα)^3]
=-1 。
答
[sin(α+pi)*cos(pi+α)*cos(α+2pi)]/[tan(pi+α)*cos^3(-α-pi)]
=[(-sin(α))*(-cos(α))*cos(α)]/[tan(α)*(-cos^3(α))]
=-sinα*cosα*cosα*cosα/sinα*cosα*cosα*cosα
=-1
应该是-1