函数f(x)=cos(-1/2)+sin(π-x/2).x∈R, ⑴求f(x)周期,⑵求f(x)在[0,π]上的减区间,⑴求f(x)周期 ,⑵求f(x)在[0,π]上的减区间(3)若f(a)=(2√10)/5,a属于(0,π/2),求tan(2a+π/4)的值重点是第三小题
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![函数f(x)=cos(-1/2)+sin(π-x/2).x∈R, ⑴求f(x)周期,⑵求f(x)在[0,π]上的减区间,⑴求f(x)周期 ,⑵求f(x)在[0,π]上的减区间(3)若f(a)=(2√10)/5,a属于(0,π/2),求tan(2a+π/4)的值重点是第三小题](/uploads/image/z/1996726-22-6.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3Dcos%28-1%2F2%29%2Bsin%28%CF%80-x%2F2%29.x%E2%88%88R%2C+%E2%91%B4%E6%B1%82f%28x%29%E5%91%A8%E6%9C%9F%2C%E2%91%B5%E6%B1%82f%28x%29%E5%9C%A8%5B0%2C%CF%80%5D%E4%B8%8A%E7%9A%84%E5%87%8F%E5%8C%BA%E9%97%B4%2C%E2%91%B4%E6%B1%82f%28x%29%E5%91%A8%E6%9C%9F+++++++%2C%E2%91%B5%E6%B1%82f%28x%29%E5%9C%A8%5B0%2C%CF%80%5D%E4%B8%8A%E7%9A%84%E5%87%8F%E5%8C%BA%E9%97%B4%EF%BC%883%EF%BC%89%E8%8B%A5f%EF%BC%88a%EF%BC%89%3D%EF%BC%882%E2%88%9A10%EF%BC%89%2F5%2Ca%E5%B1%9E%E4%BA%8E%EF%BC%880%2C%CF%80%2F2%EF%BC%89%2C%E6%B1%82tan%EF%BC%882a%2B%CF%80%2F4%EF%BC%89%E7%9A%84%E5%80%BC%E9%87%8D%E7%82%B9%E6%98%AF%E7%AC%AC%E4%B8%89%E5%B0%8F%E9%A2%98)
函数f(x)=cos(-1/2)+sin(π-x/2).x∈R, ⑴求f(x)周期,⑵求f(x)在[0,π]上的减区间,⑴求f(x)周期 ,⑵求f(x)在[0,π]上的减区间(3)若f(a)=(2√10)/5,a属于(0,π/2),求tan(2a+π/4)的值重点是第三小题
函数f(x)=cos(-1/2)+sin(π-x/2).x∈R, ⑴求f(x)周期,⑵求f(x)在[0,π]上的减区间
,⑴求f(x)周期 ,⑵求f(x)在[0,π]上的减区间
(3)若f(a)=(2√10)/5,a属于(0,π/2),求tan(2a+π/4)的值
重点是第三小题 前两个小题不写也行
函数f(x)=cos(-1/2)+sin(π-x/2).x∈R, ⑴求f(x)周期,⑵求f(x)在[0,π]上的减区间,⑴求f(x)周期 ,⑵求f(x)在[0,π]上的减区间(3)若f(a)=(2√10)/5,a属于(0,π/2),求tan(2a+π/4)的值重点是第三小题
函数f(x)=cos(-x/2)+sin(π-x/2).x∈R,⑴求f(x)周期,⑵求f(x)在[0,π]上的减区间
()求f(x)周期;(2)求f(x)在[0,π]上的减区间;
(3)若f(α)=(2√10)/5,a属于(0,π/2),求tan(2a+π/4)的值
f(x)=cos(x/2)+sin(π-x/2)=cos(x/2)+sin(x/2)=(√2)cos(x/2-π/4)
(1)最小正周期T=2π/(1/2)=4π
(2)当0≦x≦π/2时f(x)单调增;π/2≦x≦π时f(x)单调减.
(3)f(α)=cos(α/2)+sin(α/2)=(2√10)/5,
平方之得1+sinα=40/25=8/5,故sinα=3/5,cosα=4/5,sin2α=2sinαcosα=24/25;
cos2α=cos²α-sin²α=16/25-9/25=7/25;tan2α=sin2α/cos2α=24/7
于是tan(2α+π/4)=(1+tan2α)/(1-tan2α)=(1+24/7)/(1-24/7)=-31/17
函数好像写错了吧 cos里边应该有x的吧