作业帮设函数f(x)=2x-cosx,{An}是公差为(π/8)的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a3=
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![设函数f(x)=2x-cosx,{An}是公差为(π/8)的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a3=](/uploads/image/z/2507275-19-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D2x-cosx%2C%7BAn%7D%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%BA%28%CF%80%2F8%29%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2Cf%28a1%29%2Bf%28a2%29%2B%E2%80%A6f%28a5%29%3D5%CF%80%2C%E5%88%99%5Bf%28a%29%5D%5E2-a1%2Aa3%3D)
设函数f(x)=2x-cosx,{An}是公差为(π/8)的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a3=
设函数f(x)=2x-cosx,{An}是公差为(π/8)的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a3=
设函数f(x)=2x-cosx,{An}是公差为(π/8)的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a3=
因为cos(-3π/2)+cos(-π/2)+cosπ/2+cos3π/2+cos5π/2=0
2(-3π/2-π/2+π/2+3π/2+5π/2)=5π
所以a1=-3π/2
a2=-π/2
a3=π/2
a4=3π/2
a5=5π/2
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设函数f(x)=2x-cosx,{An}是公差为(π/8)的等差数列,f(a1)+f(a2)+…f(a5)=5π,则[f(a)]^2-a1*a3=
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