高一数学题已知对于任意实数x,均有f(π-x)=-f(x),f(2π-x)=f(x)成立,且当x∈(0,π/2)时,有f(x)=x^2,试求f(21π/5)的值.
2019-05-27
高一数学题
已知对于任意实数x,均有f(π-x)=-f(x),f(2π-x)=f(x)成立,且当x∈(0,π/2)时,有f(x)=x^2,试求f(21π/5)的值.
优质解答
f(21π/5)=f[2π-(-11π/5)]=f(-11π/5)=f(π-16π/5)
=-f(16π/5)=-f[2π-(-6π/5)]=-f(-6π/5)=-f(π-11π/5)
=f(11π/5)=f[2π-(-π/5)]=f(-π/5)=f(π-6π/5)
=-f(6π/5)=f(2π-4π/5)=f(4π/5)=(4π/5)^2
f(21π/5)=f[2π-(-11π/5)]=f(-11π/5)=f(π-16π/5)
=-f(16π/5)=-f[2π-(-6π/5)]=-f(-6π/5)=-f(π-11π/5)
=f(11π/5)=f[2π-(-π/5)]=f(-π/5)=f(π-6π/5)
=-f(6π/5)=f(2π-4π/5)=f(4π/5)=(4π/5)^2