已知定义域为R的函数f(x)对任意实数x,y满足f(x+y)+f(x-y)=2f(x)cosy,且f(0)=0,f(π/2)=1

证:由f(x+y)+f(x-y)=2f(x)cosy 得f(0+y)+f(0-y)=2f(0)cosy=0,化简得 f(y)+f(-y)=0亦即 f(-y)=-f(y)所以 f(x)为奇函数.由f(x+y)+f(x-y)=2f(x)cosy 得f(x+π/2)+f(x-π/2)=2f(x)cosπ/2=0,亦即f(x+π/2)+f(x-π/2)=0 ...令x=m+π，m是什么？为什么要令x=m+π？f(m+3π/2)+f(m+π/2)=0 亦即 f(x+3π/2)+f(x+π/2)=0是怎么变的？①②得 f(x+3π/2)=f(x-π/2)是怎么来的？麻烦了，谢谢！这三个问题,都是归纳为:x,m都是任意实数,不用纠结于字母,把它看成是任意实数就可以了前面得到:f(x+π/2)+f(x-π/2)=0 ① >>令 x=m+π 代入得f(m+π+π/2)+f(m+π-π/2)=0化简得f(m+3π/2)+f(m+π/2)=0③到这一步,其实 x,m都是任意实数的意思,也就是③等价于:f(x+3π/2)+f(x+π/2)=0 而:f(x+π/2)+f(x-π/2)=0 ①f(x+3π/2)+f(x+π/2)=0②f(x+π/2)+f(x-π/2)=f(x+3π/2)+f(x+π/2)=0消去f(x+π/2)所以f(x-π/2)=f(x+3π/2)