只用回答第二个问题 Thx已知函数f(x)=sin2x,g(x)=cos(2x+π/6),直线x=t(t∈R)与函数f(x)、g(x)的图象分别交于M、N两点.(1)当t=π/4时,求‖MN‖的值;(2)求‖MN‖在t∈[0,π/2]时的最大值.
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![只用回答第二个问题 Thx已知函数f(x)=sin2x,g(x)=cos(2x+π/6),直线x=t(t∈R)与函数f(x)、g(x)的图象分别交于M、N两点.(1)当t=π/4时,求‖MN‖的值;(2)求‖MN‖在t∈[0,π/2]时的最大值.](/uploads/image/z/4377831-15-1.jpg?t=%E5%8F%AA%E7%94%A8%E5%9B%9E%E7%AD%94%E7%AC%AC%E4%BA%8C%E4%B8%AA%E9%97%AE%E9%A2%98+Thx%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin2x%2Cg%28x%29%3Dcos%282x%2B%CF%80%2F6%29%2C%E7%9B%B4%E7%BA%BFx%3Dt%28t%E2%88%88R%29%E4%B8%8E%E5%87%BD%E6%95%B0f%28x%29%E3%80%81g%28x%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E5%88%86%E5%88%AB%E4%BA%A4%E4%BA%8EM%E3%80%81N%E4%B8%A4%E7%82%B9.%EF%BC%881%EF%BC%89%E5%BD%93t%3D%CF%80%2F4%E6%97%B6%2C%E6%B1%82%E2%80%96MN%E2%80%96%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E2%80%96MN%E2%80%96%E5%9C%A8t%E2%88%88%5B0%2C%CF%80%2F2%5D%E6%97%B6%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC.)
只用回答第二个问题 Thx已知函数f(x)=sin2x,g(x)=cos(2x+π/6),直线x=t(t∈R)与函数f(x)、g(x)的图象分别交于M、N两点.(1)当t=π/4时,求‖MN‖的值;(2)求‖MN‖在t∈[0,π/2]时的最大值.
只用回答第二个问题 Thx
已知函数f(x)=sin2x,g(x)=cos(2x+π/6),直线x=t(t∈R)与函数f(x)、g(x)的图象分别交于M、N两点.
(1)当t=π/4时,求‖MN‖的值;
(2)求‖MN‖在t∈[0,π/2]时的最大值.
只用回答第二个问题 Thx已知函数f(x)=sin2x,g(x)=cos(2x+π/6),直线x=t(t∈R)与函数f(x)、g(x)的图象分别交于M、N两点.(1)当t=π/4时,求‖MN‖的值;(2)求‖MN‖在t∈[0,π/2]时的最大值.
第一问 简单代入 得1+二分之根号三
第二问 将t代入 灵活运用2倍角公式
sin2t+sin(π/2-(2t+π/6)=sin2t-sin(π/3-2t)
=sin2t+sin(2t-π/3)=sin(2t-π/6+π/6)+sin(2t-π/6-π/6) (用二倍角公式)
=2sin(2t-π/6)cosπ/6
因为t∈[0,π/2] 所以:2t-π/6∈[-π/6,5π/6]
y=sinx最大值是1 同理sin(2t-π/6)有最大值1
2sin(2t-π/6)有最大值2
所以‖MN‖的最大值为│2cosπ/6│=3^(1/2)=根号3
(2)当x=t时,f(x)=sin2x=sin2t,g(x)=cos(2x+π/6)=cos(2t+π/6)
所以‖MN‖=│sin2t-cos(2t+π/6)│=│sin2t-sin[π/2-(2t+π/6)]│
=│sin2t-sin(π/3-2t)│=│sin2t+sin(2t-π/3)│
=│si...
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(2)当x=t时,f(x)=sin2x=sin2t,g(x)=cos(2x+π/6)=cos(2t+π/6)
所以‖MN‖=│sin2t-cos(2t+π/6)│=│sin2t-sin[π/2-(2t+π/6)]│
=│sin2t-sin(π/3-2t)│=│sin2t+sin(2t-π/3)│
=│sin(2t-π/6+π/6)+sin(2t-π/6-π/6)│
=│2sin(2t-π/6)cosπ/6│
因为t∈[0,π/2] 所以:2t-π/6∈[-π/6,5π/6]
当2t-π/6=π/2时,sin(2t-π/6)有最大值1,此时也是‖MN‖取最大值
所以‖MN‖的最大值为│2cosπ/6│=3^(1/2)====根号3
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