化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/14 18:26:42
化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)]
[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
=[(-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
化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(
求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan(
若α∈(0,π/6) 比较tan(sinα),tan(cosα),tan(tanα)的大小
化简{sin(π+α)^3cos(-α)cos(π+α)}/{tan(π+α)^3cos(-π-α)^2}
化简sin³(﹣α)cos(2π+α)tan(2π-α)/sin(α-2π)cos(α-3π/2)-tan(π
计算sin(-α-5π)*cos(α-2分之π)-tan(α-2分之3π)*tan(2π-α)
f(α)=[sin(π-α)cos(α-π)tan(-α+π)]/[tan(π+α)cos(3/2π+α)] 化简
化简sin(α-2π)cos(3π/2-α)+tan(π-α)tan(π/2+α)+cos^2(π-α)
化简:(1)、tan(2π+α)tan(π+α)cos(-π-α)的3次方分之sin(-α)的平方乘以cos(π+α);
tan(π-α)*sin^2(α+π/2)*cos(2π-α)/cos^3(-α-π)*tan(2π-α) sin^2就
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
(1+tanα)/(1-tanα)=3+2√2,求cos^2(π-α)+sin(π+α)*cos(π-α)+2sin(α