求y=sin(π/2-x)的单调递增区间?判断奇偶性f(x)=√2sin2x f(x)=sin(3x/4+3π/2) f(x)=√(1-cosx) + √(cosx-1)
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![求y=sin(π/2-x)的单调递增区间?判断奇偶性f(x)=√2sin2x f(x)=sin(3x/4+3π/2) f(x)=√(1-cosx) + √(cosx-1)](/uploads/image/z/4336692-60-2.jpg?t=%E6%B1%82y%3Dsin%28%CF%80%2F2-x%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%3F%E5%88%A4%E6%96%AD%E5%A5%87%E5%81%B6%E6%80%A7f%28x%29%3D%E2%88%9A2sin2x++++++++++++++++++f%28x%29%3Dsin%283x%2F4%2B3%CF%80%2F2%29++++++++++++++f%28x%29%3D%E2%88%9A%281-cosx%29++%2B+%E2%88%9A%28cosx-1%29)
求y=sin(π/2-x)的单调递增区间?判断奇偶性f(x)=√2sin2x f(x)=sin(3x/4+3π/2) f(x)=√(1-cosx) + √(cosx-1)
求y=sin(π/2-x)的单调递增区间?
判断奇偶性f(x)=√2sin2x f(x)=sin(3x/4+3π/2) f(x)=√(1-cosx) + √(cosx-1)
求y=sin(π/2-x)的单调递增区间?判断奇偶性f(x)=√2sin2x f(x)=sin(3x/4+3π/2) f(x)=√(1-cosx) + √(cosx-1)
y=sin(π/2-X)=cosx
cosx的单调增区间为-π+2kπ≤x≤2kπ
所以y=sin(π/2-X)单调增区间为-π+2kπ≤x≤2kπ
f(-x)=√2sin[2(-x)]=-√2 sin2x=-f(x)
所以是奇函数
f(x) = sin(3x/4 + 3π/2)
= -cos(3x/4)
f(-x) = -cos(3(-x)/4) = -cos(-(3x/4)) = -cos(3x/4) = f(x)
∴f(x)是偶函数
f(x)=√(1-cosx)+√(cosx-1)
因为 1-cosx≥0 cosx-1≥0
所以 1-cosx=0
得 cosx=1
f(x)=0
所以函数为奇函数也是偶函数