f(x)=sin[πx-(π/2)]-1如何化简成COS(π x)-1?
f(x)=sin[πx-(π/2)]-1如何化简成COS(π x)-1?
已知f(x)=cos^2x/1+sin^2x求f'(π/4)
求下列函数的奇偶性f(x)=(1+sin x-cos x)/(1+cos x+sin x),x属于[(-π/2),(π/
已知f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+cos平方x.
已知函数f(x)=(1+1/tanx)sin(x)^2 -2sin(x+π/4)*cos(x+π/4)
函数f(x)=-√2(sin2x+π/4)+6 sin x cos x-2cos²x+1
已知函数f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1
化简f(x)=(1+√2*cos(2x-π/4))/sin(π/2-x)
设函数f(x)=cos(2x+π/3)+sin^2x-1/2
已知函数f(x)=sin²(π/4+x)+cos²x+1/2求最值
已知函数f(x)=cos(2x-π/3)+sin^2 x-cos^2 x
已知函数f(x)=cos(2x-π\3)+sin²x-cos²x