已知tanα=-1/3,cosβ=(根号5)/5,α,β∈(0,π).(1)求tan(α+β)的值(2)求函数f(x)=(√2)sin(x-α)+cos(x+β)的最大值
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![已知tanα=-1/3,cosβ=(根号5)/5,α,β∈(0,π).(1)求tan(α+β)的值(2)求函数f(x)=(√2)sin(x-α)+cos(x+β)的最大值](/uploads/image/z/3712142-38-2.jpg?t=%E5%B7%B2%E7%9F%A5tan%CE%B1%3D-1%2F3%2Ccos%CE%B2%3D%28%E6%A0%B9%E5%8F%B75%29%2F5%2C%CE%B1%2C%CE%B2%E2%88%88%280%2C%CF%80%29.%281%29%E6%B1%82tan%28%CE%B1%2B%CE%B2%29%E7%9A%84%E5%80%BC%282%29%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%3D%28%E2%88%9A2%29sin%28x-%CE%B1%29%2Bcos%28x%2B%CE%B2%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC)
已知tanα=-1/3,cosβ=(根号5)/5,α,β∈(0,π).(1)求tan(α+β)的值(2)求函数f(x)=(√2)sin(x-α)+cos(x+β)的最大值
已知tanα=-1/3,cosβ=(根号5)/5,α,β∈(0,π).
(1)求tan(α+β)的值
(2)求函数f(x)=(√2)sin(x-α)+cos(x+β)的最大值
已知tanα=-1/3,cosβ=(根号5)/5,α,β∈(0,π).(1)求tan(α+β)的值(2)求函数f(x)=(√2)sin(x-α)+cos(x+β)的最大值
这题不难,看下图:
(1)因为α,β∈(0,π),所以sinβ=2(根号5)/5,所以tanβ=2,所以tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=1
(2)tanα=-1/3,tanβ=2,tan(α+β)=1,α,β∈(0,π),所以α+β=5π/4.
f(x)=(√2)sin(x-α)+cos(x+β)
=(√2)sin(x-α)+cos(x-α+5π/4)
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(1)因为α,β∈(0,π),所以sinβ=2(根号5)/5,所以tanβ=2,所以tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=1
(2)tanα=-1/3,tanβ=2,tan(α+β)=1,α,β∈(0,π),所以α+β=5π/4.
f(x)=(√2)sin(x-α)+cos(x+β)
=(√2)sin(x-α)+cos(x-α+5π/4)
=(√2)sin(x-α)-cos(x-α+π/4)
=(√2)sin(x-α)-(√2)/2cos(x-α)+(√2)/2sin(x-α)
=(3√2)/2sin(x-α)-(√2)/2cos(x-α)
=√5cos(x-α-c) 其中,tan(c)=3
所以最大值为√5.
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