已知cos(x+π/6)=5/13,x∈(0,π/2),则cosx=
来源:学生作业帮助网 编辑:六六作业网 时间:2025/02/02 19:28:52
已知cos(x+π/6)=5/13,x∈(0,π/2),则cosx=已知cos(x+π/6)=5/13,x∈(0,π/2),则cosx=已知cos(x+π/6)=5/13,x∈(0,π/2),则cos
已知cos(x+π/6)=5/13,x∈(0,π/2),则cosx=
已知cos(x+π/6)=5/13,x∈(0,π/2),则cosx=
已知cos(x+π/6)=5/13,x∈(0,π/2),则cosx=
x∈(0,π/2)
x+π/6∈(π/6,2π/3)
所以sin(x+π/6)>0
sin²(x+π/6)+cos²(x+π/6)=1
所以sin(x+π/6)=12/13
cosa
=cos[(x+π/6)-π/6]
=cos(x+π/6)cosπ/6+sin(x+π/6)sinπ/6
=(5√3+12)/26
已知cos(x+π/12)=-5/13,x是第三象限角,求cos(x-π/6)的值
已知cos(x+π/12)=-5/13,x是第三象限角,求cos(x-π/6)的值
已知函数f(x)=cos^2x+sinxcosx,x∈Rf(π/6)
已知cos(x+π/6)=5/13,x∈(0,π/2),则cosx=
已知sinx=3/5,x∈(π/2,π),求【sin(x+y)+sin(x-y)】/【cos(x+y)+cos(x-y)】的值
已知cos(13π/6+x)=根3/3,求cos(23π/6-x)的值
已知cos(x+派/6)=5/13,0
已知函数f(x)=根2cos(x-x/12),x∈R.求f(-π/6)的值.2)若cosy=3/5
已知函数f(x)=√2cos(x-π/12),x∈R
已知函数f(x)=[2sin(x-π/6)+√3sin x]cos x+sin^2x,x∈R
已知已知cos(π/4+x)=5/13,且0
已知0《x《π/4,sin(π/4-x)=5/13,求cos2x/cos(π/4+x)
已知sinx+cosx=1/3,则sinxcosx=已知tanx=2,则sinx+cosx/sinx-cosx= sin^2x-cos^2x+sinxcosx=化简2sin^2x-1/1-2cos^2x=若sin(π/6-x)=a,则cos(2π/3-x)=已知sinα=3/5,α∈(π/2,π) cosβ=5/13,β∈(3π/2,2π),求cos(α-β)=
已知cos(x-π/6)=-根号三分之三 则cos(x)+cos(x-π/3)的值是
已知cos(π/6+x)=√3/3,求cos(5π/6-x)的值,
已知cos(x-π╱6)=m,则cosx+cos(x-π╱3)=
求值:sin(π/4+3x)cos(3/π-3x)+cos(π/6+3x)cos(π/4+3x)其中:π=派2.已知sinα=3/5,cosβ=15/17,求cos(α-β)的值3.已知tanθ=4/3,θ∈(0,π/2),求cos(2π/3-θ)
已知函数f(x)=sin(2x+π/6)+sin(2x+π/6)+2cos²x