已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/07 04:36:10
已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?
已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于() A.-2009 B.-1/2009 C.1/2009 D.2009
已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于() A.-2009 B.-1/2009 C.1/2009 D.2009
【参考答案】D
[sin(π/2-x)+sin(π-x)]/[cos(-x)+sin(2π-x)]=2009
根据诱导公式,化简
(cosx+sinx)/(cosx-sinx)=2009
左边分子分母同时除以cosx:
(1+tanx)/(1-tanx)=2009
∵tan5π/4=tan(π+π/4)=1
∴(tan5π/4+tanx)/(1-tan5π/4tanx)=2009
根据两角和正切公式
tan(x+5π/4)=(tan5π/4+tanx)/(1-tan5π/4tanx)=2009
∴tan(x+5π/4)=2009
欢迎追问...
[sin(π/2-x)+sin(π-x)]/[cos(-x)+sin(2π-x)]=2009
根据诱导公式,化简
(cosx+sinx)/(cosx-sinx)=2009
左边分子分母同时除以cosx:
(1+tanx)/(1-tanx)=2009
∵tan5π/4=tan(π+π/4)=1
∴(tan5π/4+tanx)/(1-tan5π/4tanx)=2009
根据两角和正切公式
tan(x+5π/4)=(tan5π/4+tanx)/(1-tan5π/4tanx)=2009
∴tan(x+5π/4)=2009
欢迎追问...
已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?
已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x
已知(x)=[sin(π-x)cos(2π-x)tan(-x+π)]/[cos(-π/2+x)]
化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π
三角函数题.已知tanx=2.求(sin(π-x)cos(2π-x)sin(-x+3π/2))/tan(-x-π)sin
已知f(x)=cos^2x+sinxcosx g(x)=2sin(x+π/4)sin(x-π/4)
.已知sin x+cos x= (0≤x<π),则tan x的值等于
已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
已知函数f(x)= [sin(2π-x)sin(π+x)cos(-x-π)] /[2cos(π-x)sin(3π-x)]
已知f(x)=(sin(π-x)*cos(2π-x))/cos(-π-x)*tan(π-x),则f(-31π/3)的值为
已知函数f(x)=(1+1/tanx)sin(x)^2 -2sin(x+π/4)*cos(x+π/4)
已知函数f(x)=sin^2*x-根号3*sinπ/4*x*cosπ/4*x