求证cosx+sinx=√2cos(x-π/4)

求证:tan(x-π/4)=(sinx-cosx)/(sinx+cosx) 求证sinx-cosx=根号2sin(x-π/4) 已知tan(x+(5π)/4)=2,求(sinx+cosx)/(sinx-cosx)+sin2x+cos(x)^2 已知函数f(x)=cos(2x-π/3)+(cosx+sinx)(cosx-sinx) sinx+cosx/sinx-cosx=2 求sinx/cos^3x +cosx/sin^3x 1-2sinx cosx /COS^2X-SIN^2X =1-tanx/1+tanx 求证 已知向量a=(2cosx,sinx)向量b={cos(x-π/3),√3cosx-sinx} 求证 sin^2x/(sinx-cosx)-(sinx+cosx)/tan^2 x-1=sinx+cosx 已知向量a=(4sinx,-4√6(cosx+sinx),b=(2cosx,cos(x+π/4),f(x)=a·b【a, 根号2cos(x-π/2)怎么=sinx+cosx cosx+根号3sinx=2cos(x-π/3) 已知sinx+cosx/sinx-cosx=2,求2cos(π-x)-3sin(π+x)/4cos(-x)+sin(2π