f(x)=cosx+sinx,證明存在a屬於(0,0.5π)使f(x+a)=f(x+3a)恒成立

f(x)=cosx+sinx,證明存在a屬於(0,0.5π)使f(x+a)=f(x+3a)恒成立

f(x+a)=f(x+3a)sin(x+a)+cos(x+a)=sin(x+3a)+cos(x+3a)sin(x+a)-sin(x+3a)=cos(x+3a)-cos(x+a)2cos(x+2a)sin(-a)=-2sin(x+2a)sin(a)sina[cos(x+2a)-sin(x+2a)]= 0sina=0,因a屬於(0,π/2),則不存…