Question

化简：[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)]

Answer 1

[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)]
=[(-sinα)*(-cosα)*(-tgα)]/[(-sinα)*ctgα*(tgα)]
=[-sinα*cosα*tgα]/[-sinα*ctgα*tgα]
=cosα/ctgα
=cosαtgα

诱导公式：
sin（2kπ＋α）＝sinα
cos（2kπ＋α）＝cosα
tg（2kπ＋α）＝tgα
ctg（2kπ＋α）＝ctgα

sin（－α）＝－sinα
cos（－α）＝cosα
tg（－α）＝－tgα
ctg（－α）＝－ctgα

sin（π＋α）＝－sinα
cos（π＋α）＝－cosα
tg（π＋α）＝tgα
ctg（π＋α）＝ctgα

sin（π－α）＝sinα
cos（π－α）＝－cosα
tg（π－α）＝－tgα
ctg（π－α）＝－ctgα

sin（π/2＋α）＝cosα
cos（π/2＋α）＝-sinα
tg（π/2＋α）＝－ctgα
ctg（π/2＋α）＝－tgα

sin（π/2－α）＝cosα
cos（π/2－α）＝sinα
tg（π/2－α）＝ctgα
ctg（π/2－α）＝tgα

Answer 2

解：：[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)] =
：[-sinα*(-cosα)*(-tanα)]/[(-sinα)*tanα*tanα]
=cosα*cotα