[f(a)]^2-a1*a3=
中是f(a?)
由已知,f(a1)+f(a2)+...+f(a5)=5π
带入得10a3-cos(a3-π/4)-cos(a3-π/8)-cosa3-cos(a3+π/8)-cos(a3+π/4)=5π
化简得10a3-2cosa3cosπ/8-2cosa3cosπ/4-cosa3=5π
令g（a3)=10a3-2cosa3cosπ/8-2cosa3cosπ/4-cosa3=5π
则g’(a3)(它的导函数）=10+（2cosπ/8+2cosπ/4+1)sina3＞0
所以g(a3)单增
而a3=π/2显然是其一个解
所以g(a3)=5π有且只有a3=π/2一个解
易得[f(a3)]^2-a1×a3=π^2-π/4×π/2=7π^2/8