定义在R上的函数f(x)满足①f(x+y)+f(x-y)=2f(x)cosy②f(0)=0,f(π/2)=1.求f(x)

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定义在R上的函数f(x)满足①f(x+y)+f(x-y)=2f(x)cosy②f(0)=0,f(π/2)=1.求f(x)
我已经证得f(x)为奇函数不知道怎么赋值求f(x) P.S.请勿复制 Thx.

令y=π/2,得f(x+π/2)+f(x-π/2)=2f(x)cosπ/2=0,
所以f(x+π/2)=-f(x-π/2)=f(π/2-x)
令x=π/2,得f(π/2+y)+f(π/2-y)=2f(π/2)cosy=2cosy,
f(π/2+y)+f(y+π/2)=2cosy
所以f(y+π/2）=cosy,
再令y+π/2=x,
得f(x)=sinx

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