已知函数f(x)=cos^2(x+π/12)+1/2sin2x求f(x)的最值求f(x)的单调增区间另外,想问这个变换是怎么来的,为什么2乘进(x+π/12)之后,整个cos还要除于2:cos^2(x+π/12)=[cos(x+(π/12)]^2=[cos(2x+(π/6))+1]/2
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![已知函数f(x)=cos^2(x+π/12)+1/2sin2x求f(x)的最值求f(x)的单调增区间另外,想问这个变换是怎么来的,为什么2乘进(x+π/12)之后,整个cos还要除于2:cos^2(x+π/12)=[cos(x+(π/12)]^2=[cos(2x+(π/6))+1]/2](/uploads/image/z/5440853-29-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dcos%5E2%28x%2B%CF%80%2F12%29%2B1%2F2sin2x%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E6%9C%80%E5%80%BC%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%8D%95%E8%B0%83%E5%A2%9E%E5%8C%BA%E9%97%B4%E5%8F%A6%E5%A4%96%2C%E6%83%B3%E9%97%AE%E8%BF%99%E4%B8%AA%E5%8F%98%E6%8D%A2%E6%98%AF%E6%80%8E%E4%B9%88%E6%9D%A5%E7%9A%84%2C%E4%B8%BA%E4%BB%80%E4%B9%882%E4%B9%98%E8%BF%9B%EF%BC%88x%2B%CF%80%2F12%EF%BC%89%E4%B9%8B%E5%90%8E%2C%E6%95%B4%E4%B8%AAcos%E8%BF%98%E8%A6%81%E9%99%A4%E4%BA%8E2%EF%BC%9Acos%5E2%28x%2B%CF%80%2F12%29%3D%5Bcos%28x%2B%28%CF%80%2F12%29%5D%5E2%3D%5Bcos%282x%2B%28%CF%80%2F6%29%29%2B1%5D%2F2)
已知函数f(x)=cos^2(x+π/12)+1/2sin2x求f(x)的最值求f(x)的单调增区间另外,想问这个变换是怎么来的,为什么2乘进(x+π/12)之后,整个cos还要除于2:cos^2(x+π/12)=[cos(x+(π/12)]^2=[cos(2x+(π/6))+1]/2
已知函数f(x)=cos^2(x+π/12)+1/2sin2x
求f(x)的最值
求f(x)的单调增区间
另外,想问这个变换是怎么来的,为什么2乘进(x+π/12)之后,整个cos还要除于2:cos^2(x+π/12)=[cos(x+(π/12)]^2=[cos(2x+(π/6))+1]/2
已知函数f(x)=cos^2(x+π/12)+1/2sin2x求f(x)的最值求f(x)的单调增区间另外,想问这个变换是怎么来的,为什么2乘进(x+π/12)之后,整个cos还要除于2:cos^2(x+π/12)=[cos(x+(π/12)]^2=[cos(2x+(π/6))+1]/2
这是利用公式 cos2x=2(cosx)^2-1 (cosx)^2=1/2(cos2x+1)
f(x) = 1/2 cos(2x+pi/6), g(x) = 1+ 1/2 sin2x
f(x) 对称轴为2x+pi/6 = 0, 2pi, ...
x = -pi/12 +/- npi...
f(x)+g(x) = 1/2 cos2x * sqrt(3)/2 - 1/2 sin2x * 1/2 + 1 + 1/2 sin2x
= 1/2 cos2x sqrt(3...
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f(x) = 1/2 cos(2x+pi/6), g(x) = 1+ 1/2 sin2x
f(x) 对称轴为2x+pi/6 = 0, 2pi, ...
x = -pi/12 +/- npi...
f(x)+g(x) = 1/2 cos2x * sqrt(3)/2 - 1/2 sin2x * 1/2 + 1 + 1/2 sin2x
= 1/2 cos2x sqrt(3)/2 + 1/2 sin2x 1/2 + 1 = 1/2 cos(2x-pi/6) + 1
最小正周期派,值域[-1/2, 3/2]
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