若cos(π/4+x)=3/5,17/12π<x<7/4π,求(sin2x+2sinx)/(1-tanx)的值
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已知cos(x+45º)=3/5且17π/12<x<7π/4,求(sin2x+2cos²x)/1-tanx的值注意!括号里加的是cos^2x
cos(π/4+x)=3/5,17π/12
cos(π/4+X)=3/5,17π/12
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17/12π
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,17π/12
cos(π/4+x)=3/5,-17π/12
已知sin[a-b]cos a-cos[b-a]sin a=3/5,b是第三象限角,求sin[b+5π/4]的值第一题1/2cos x-√3/2sin x第二题√3sin x+cos x第三题√2[sin x-cos x]第四题√2cos x-√6sin x
若cos(x+y)=4/5,cos(x-y)=-4/5,且3π/2
三角函数解方程 (cos 3x)^3+ (cos 5x)^3 = 8 (cos 4x)^3 (cos x)^3
化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π)
化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π)
若cos(π/4+x)=3/5,17/12π<x<7/4π,求(sin2x+2sinx)/(1-tanx)的值
cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0
函数y=sin(2分之π+x)cos(6分之π-x)的最大值是我看了很多人的解答 他们做的都对 可我就是不理解 比如说这个解答原式=cos(x)*cos(π/6-x)=1/2*[cos(π/6)+cos(2x-π/6)]=1/2*cos(2x-π/6)+√3/4Y=sin(派/2+x)cos(派/6-x
设cos (x +π/4 )=3/5,17π/12