已知函数f(x)满足f(x+y)+f(x-y)=2f(x)•f(y),且f(0)≠0,若f(π2
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已知函数f(x)满足f(x+y)+f(x-y)=2f(x)•f(y),且f(0)≠0,若f(
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![已知函数f(x)满足f(x+y)+f(x-y)=2f(x)•f(y),且f(0)≠0,若f(π2](/uploads/image/z/18708360-24-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E6%BB%A1%E8%B6%B3f%EF%BC%88x%2By%EF%BC%89%2Bf%EF%BC%88x-y%EF%BC%89%3D2f%EF%BC%88x%EF%BC%89%E2%80%A2f%EF%BC%88y%EF%BC%89%EF%BC%8C%E4%B8%94f%EF%BC%880%EF%BC%89%E2%89%A00%EF%BC%8C%E8%8B%A5f%EF%BC%88%CF%802)
令x=y=0得f(0)+f(0)=2f(0)×f(0)=2f(0)
而f(0)≠0∴f(0)=1
令x=y=
π
2得f(π)+f(0)=2f(
π
2)f(
π
2)=0
∴f(π)=-1
令x=y=π,得f(2π)+f(0)=2f(π)f(π)=2
∴f(2π)=1
而f(0)≠0∴f(0)=1
令x=y=
π
2得f(π)+f(0)=2f(
π
2)f(
π
2)=0
∴f(π)=-1
令x=y=π,得f(2π)+f(0)=2f(π)f(π)=2
∴f(2π)=1
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