化简(1)sin(α+π)cos(-α)sin(-α-π) (2)sin³(-α)cos(2π+α)tan(-α-π)(1)sin(α+π)cos(-α)sin(-α-π) (2)sin³(-α)cos(2π+α)tan(-α-π)

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/08 23:11:22

化简(1)sin(α+π)cos(-α)sin(-α-π) (2)sin³(-α)cos(2π+α)tan(-α-π)(1)sin(α+π)cos(-α)sin(-α-π) (2)sin³(-α)cos(2π+α)tan(-α-π)
化简(1)sin(α+π)cos(-α)sin(-α-π) (2)sin³(-α)cos(2π+α)tan(-α-π)
(1)sin(α+π)cos(-α)sin(-α-π)
(2)sin³(-α)cos(2π+α)tan(-α-π)

化简(1)sin(α+π)cos(-α)sin(-α-π) (2)sin³(-α)cos(2π+α)tan(-α-π)(1)sin(α+π)cos(-α)sin(-α-π) (2)sin³(-α)cos(2π+α)tan(-α-π)
(1)sin(α+π)cos(-α)sin(-α-π)
=(-sinα)*cosα*(sinα)
=-sin²α*cosα
(2)sin³(-α)cos(2π+α)tan(-α-π)
=(-sinα)³*cosα*(-tanα)
=-sin³α*cosα*(-sinα/cosα)
=(sinα)^4

(1)-sin(α)^2cos(α)
(2)sin(α)^4

(1)sin2αcosα (2)sin4α

化简(1-sinα+cosα/1-sinα-cosα )+(1-sinα-cosα/1-sinα+cosα) (1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 (3/2*π 若α∈(0,π),化简[1+sinα+cosα)(sinα/2-cosα/2)]/根号(2+2cosα) 化简:tanα(cosα-sinα)+sinα(sinα+tanα)/1+cosα. 设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π+α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2(-sinα)(-cosα)+cosα/[1+sin² 已知6sin²α+sinαcosα-2cos²α=0,α∈(π/2,π),求值1).(sinα-3cosα)/(sinα-cosα);2).sinαcosα-sin²α;3).sin²α-3cosαsinα-2 sin(α-β)sin(β-r)-cos(α-β)cos(r-β) 1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)2.( tan4分之5π+tan12分之5π)/(1-tan12分之5π)3.[ sin(α+β)-2sinαcosβ]/2sinαsinβ+cos(α+β) 化简[sin(π/2-α)cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π+α) 求证cosα+1-sinα/cosα+1+sinα=1+sinα/cosα(cosα+1-sinα)/(cosα+1+sinα)=(1+sinα)/cosα 求证 1+sinα+cosα+2sinαcosα/求证 (1+sinα+cosα+2sinαcosα)/(1+sinα+cosα)=sinα+cosα α为第二象限角,化简cosα根号(1-sinα/1+sinα)+sinα根号(1-cosα/1+cosα) 化简 根号2cosα-根号2sinα化简(1) 根号2cosα-根号2sinα(2)sinα+cosα 已知α∈(π,2π) sinα+cosα=1/5 求sinα*cosα sinα-cosα 化简(sin²α/1-cosα)-cosα 化简cosα*根号(1-sinα)/(1+sinα)+sinα*根号(1-cosα)/(1+cosα) 已知α=7/12π,那么cosα√(1-sinα)/(1+sinα)+sinα√(1-cosα)/(1-cosα)= 已知tan(3π+α)=2,求:1、(sinα+cosα)²;2、sinα-cosα/2sinα+cosα 已知tan(π-α)=2,求sin²α-2sinαcosα-cos²α/4cos²α-3sin²α+1