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题目设函数f(x)对所有的实数x都满足f(x+2π)=f(x),求证:存在4个函数fi(x)(i=1,2,3,4)满足:(1)对i=1,2,3,4,fi(x)是偶函数,且对任意的实数x,有fi(x+π)=fi(x);(2)对任意的实数x,有f(x)=f1(x)+f2(x)cosx+f3(x)sinx+f4(x)sin2x.(2007年全国高中数学联赛一试第15题)推广设函数f(x)
The title assignment function f(x) satisfies f(x+2π)=f(x) for all real numbers x. It is verified that there are four functions fi(x) (i=1, 2, 3, 4) satisfy: 1) For i=1,2,3,4,fi(x) is an even function, and for any real number x, there is fi(x+π)=fi(x); (2) for any real number x, There f(x) = f1 (x) + f2 (x) cosx + f3 (x) sinx + f4 (x) sin2x. (National Senior High School Mathematical League Tournament Question 15 in 2007) Promotion function f(x)