设F(X)=sin(πx/4-π/6)-2cos(πx/8)2+1求最小正周期

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设F(X)=sin(πx/4-π/6)-2cos(πx/8)2+1求最小正周期
设F(X)=sin(πx/4-π/6)-2cos(πx/8)2+1求最小正周期

设F(X)=sin(πx/4-π/6)-2cos(πx/8)2+1求最小正周期
f(x)=sin(πx/4-π/6)-[2cos²(πx/8)-1]
=sin(πx/4-π/6)-cos(πx/4)
=sin(πx/4)cos(π/6)-cos(πx/4)sin(π/6)-cos(πx/4)
=根号3/2sin(πx/4)-（1/2）cos(πx/4)-cos(πx/4)
=根号3/2sin(πx/4)-（3/2）cos(πx/4）
=根号3[1/2sin(πx/4)-（根号3/2）cos(πx/4]
=根号3sin(πx/4-π/3)
=根号3sin(π/4(x-4/3))
T=2π/(π/4)=8,

是F(X)=sin(πx/4-π/6)-2cos(πx/8)^2+1还是F(X)=sin(πx/4-π/6)-2cos^2(πx/8)+1?

设f(x)={sinπx(x 设f(x)={sinπx(x 设 f(x)= {sinπx(x 设f(x)={sinπx(x 设f(x)=sin(x+4/π),则f(2x)= 设函数 f(x)=sin(2x+y),(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f x=SIN(2X+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+ φ)(-π 设函数f(x)=sin(2x+φ)(-π 设函数f(x)=sin(2x+ φ)(-π 设函数f(x)=sin(2x+φ)(-π 设f(cos^2x)=sin^2x,则f'(π/4)等于,