sin(3π+α )=1/4,求cos(π+α)/cosα[cos(π+α)-1]+cos(α-2π)/cos(α+2π
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:综合作业 时间:2024/07/24 00:40:45
sin(3π+α )=1/4,求cos(π+α)/cosα[cos(π+α)-1]+cos(α-2π)/cos(α+2π)cos(π+α)+coa(-α)
![sin(3π+α )=1/4,求cos(π+α)/cosα[cos(π+α)-1]+cos(α-2π)/cos(α+2π](/uploads/image/z/19794646-46-6.jpg?t=sin%283%CF%80%2B%CE%B1+%29%3D1%2F4%2C%E6%B1%82cos%28%CF%80%2B%CE%B1%29%2Fcos%CE%B1%5Bcos%28%CF%80%2B%CE%B1%29-1%5D%2Bcos%28%CE%B1-2%CF%80%29%2Fcos%28%CE%B1%2B2%CF%80)
解 sin(3π+α )=1/4 sin α =-1/4 cos α=±√15/4
cos(π+α)/cosα[cos(π+α)-1]+cos(α-2π)/cos(α+2π)cos(π+α)+cos(-α)
=-cosα/cosα[-cos α -1]+cosα/cosα(-cosα)+cos α
=cosα/cosα[ cos α +1]+(-cosα)+coa α
=[ cos α +1]+2cos α
=3coa α+1=±√15/4+1
cos(π+α)/cosα[cos(π+α)-1]+cos(α-2π)/cos(α+2π)cos(π+α)+cos(-α)
=-cosα/cosα[-cos α -1]+cosα/cosα(-cosα)+cos α
=cosα/cosα[ cos α +1]+(-cosα)+coa α
=[ cos α +1]+2cos α
=3coa α+1=±√15/4+1
(1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 (3/2*π
已知 sinα+2cos(5π/2+α)/cos(π-α)-sin(π/2-α)=-1/4 求(sinα+cosα)平方
已知tan(3π+α)=2,求:1、(sinα+cosα)²;2、sinα-cosα/2sinα+cosα
已知0<α<π/2,且3sinα=4cosα求(sin^2α+2sinαcosα)/(3cos^2α-1)求cos^2α
已知(sinα-cosα)/(sinα+cosα)=1/3,则cos^4(π/3+α)-cos^4(π/6-α)的值为
已知cos(π\3-2)=1\3,则cosα+根号3sinα=
已知α∈(0,π/2),且2sinα-sinαcosα-3cosα=0.求[sin(α+π/4)]/[sin2α+cos
已知α∈(π,2π) sinα+cosα=1/5 求sinα*cosα sinα-cosα
已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值
已知sinα+cosα=三分之根号二,sinα-cosα=-4/3,且α∈(-π/2,0),计算(1+sin2α+cos
已知tan(π+α)=1/3,求cos²(π+α)+sin(π+α)cos(π-α)-2cos²(3
设f(α)=2sinαcosα+cosα/1+sin²α+cos(3π/2+α)-sin²(π/2+