已知f(x)=cos(2x-π/6)+cos(2x-5π/6)-2cos^2x+1(1)求f(x)最小正周期(2)求f(x)在【-π/4,π/4】上的最大值和最小值
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![已知f(x)=cos(2x-π/6)+cos(2x-5π/6)-2cos^2x+1(1)求f(x)最小正周期(2)求f(x)在【-π/4,π/4】上的最大值和最小值](/uploads/image/z/4339692-36-2.jpg?t=%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89%3Dcos%282x-%CF%80%2F6%29%2Bcos%282x-5%CF%80%2F6%29-2cos%5E2x%2B1%EF%BC%881%EF%BC%89%E6%B1%82f%EF%BC%88x%EF%BC%89%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%EF%BC%882%EF%BC%89%E6%B1%82f%EF%BC%88x%EF%BC%89%E5%9C%A8%E3%80%90-%CF%80%2F4%2C%CF%80%2F4%E3%80%91%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC)
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已知f(x)=cos(2x-π/6)+cos(2x-5π/6)-2cos^2x+1(1)求f(x)最小正周期(2)求f(x)在【-π/4,π/4】上的最大值和最小值
已知f(x)=cos(2x-π/6)+cos(2x-5π/6)-2cos^2x+1
(1)求f(x)最小正周期(2)求f(x)在【-π/4,π/4】上的最大值和最小值
已知f(x)=cos(2x-π/6)+cos(2x-5π/6)-2cos^2x+1(1)求f(x)最小正周期(2)求f(x)在【-π/4,π/4】上的最大值和最小值
答:
f(x)=cos(2x-π/6)+cos(2x-5π/6)-2(cosx)^2+1
=2cos(2x-π/2)cos(π/3)-cos2x
=-sin2x-cos2x
=-√2sin(2x+π/4)
1)f(x)的最小正周期T=2π/2=π
2)
-π/4