由f(x)=f(x+2),
显然f(x)是以2为周期的周期函数,
又x∈[3,5]时,f(x)=2-|x-4|
故当x∈[2n-1,2n+1]时,f(x)=2-|x-2n|
显然cos2与sin2均∈[-1,1],此时n=0
故f(cos2)=2-|cos2|,f(sin2)=2-|sin2|
又2∈[0.5π,0.75π],故|sin2|>|cos2|
则2-|cos2|>2-|sin2|
故f(cos2)>f(sin2)
由f(x)=f(x+2),
显然f(x)是以2为周期的周期函数,
又x∈[3,5]时,f(x)=2-|x-4|
故当x∈[2n-1,2n+1]时,f(x)=2-|x-2n|
显然cos2与sin2均∈[-1,1],此时n=0
故f(cos2)=2-|cos2|,f(sin2)=2-|sin2|
又2∈[0.5π,0.75π],故|sin2|>|cos2|
则2-|cos2|>2-|sin2|
故f(cos2)>f(sin2)