若cos(π/4+x)=3/5 17π/12

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若cos(π/4+x)=3/517π/12若cos(π/4+x)=3/517π/12若cos(π/4+x)=3/517π/1217π/12＜x＜7π/4,得5π/3＜x+π/4＜2πcos(x-π/4

若cos(π/4+x)=3/5 17π/12
若cos(π/4+x)=3/5 17π/12

若cos(π/4+x)=3/5 17π/12
17π/12＜x＜7π/4,得5π/3＜x+π/4＜2π
cos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-sin²(x+π/4)]=-√[1-(3/5)²]=-4/5
sin(2x)=-cos(2x+π/2)=-cos[2(x+π/4)]=1-2cos²(x+π/4)=1-2•(3/5)²=7/25
[sin(2x)+2sin²x]/(1-tanx)
=2(sinxcosx+sin²x)/(1-sinx/cosx)
=2(cosx+sinx)/(1/sinx-1/cosx)
=2(cosx+sinx)sinxcosx/(cosx-sinx)
=cos(x-π/4)sin(2x)/cos(x+π/4)
=-4/5•7/25/(3/5)
=-28/75

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 若cos(x+y)=4/5,cos(x-y)=-4/5,且3π/2 若cosx=2/3,x是第四象限角,求sin(x-2π)+sin(-x-3π)cos(x-3π)/cos(π-x)-cos(-π-x)cos(x-4π)的值 三角函数解方程 (cos 3x)^3+ (cos 5x)^3 = 8 (cos 4x)^3 (cos x)^3 若cos(π/4+x)=3/5 17π/12 若cos（π/4+x）=3/5,17/12π 若cos(π/4+x)=3/5,17π/12 若cos(π/4+x)=3/5,17π/12 若cos(π/4+x)=3/5 17π/12 若cos(π/4+x)=3/5 17π/12 若cos（π/4+x)=3/5,17/12π