若cos(π/4+x)＝3/5,17/12π＜x＜7/4π,求(sin2x+2sinx)/(1-tanx)的值

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