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

化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)]
53574004 1年前 已收到2个回答 举报

爱家的男人 花朵

共回答了15个问题采纳率:86.7% 举报

[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

1年前

8

青草薇熏 幼苗

共回答了210个问题 举报

:[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

1年前

1
可能相似的问题
Copyright © 2024 YULUCN.COM - 雨露学习互助 - 18 q. 3.582 s. - webmaster@yulucn.com