函数f(x)=sin(2x-π/4)-2√2(sin^2x)的最小正周期是?
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![函数f(x)=sin(2x-π/4)-2√2(sin^2x)的最小正周期是?](/uploads/image/z/5352503-23-3.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3Dsin%282x-%CF%80%2F4%29-2%E2%88%9A2%EF%BC%88sin%5E2x%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E6%98%AF%3F)
函数f(x)=sin(2x-π/4)-2√2(sin^2x)的最小正周期是?
函数f(x)=sin(2x-π/4)-2√2(sin^2x)的最小正周期是?
函数f(x)=sin(2x-π/4)-2√2(sin^2x)的最小正周期是?
f(x)=sin2xcosπ/4-cos2xsinπ/4-2√2(1-cos2x)/2
=√2/2*sin2x-√2/2*cos2x+√2cos2x-√2
=√2/2*sin2x+√2/2*cos2x-√2
=sin2xcosπ/4+cos2xsinπ/4-√2
=sin(2x+π/4)-√2
T=2π/2=π
f(x)=sin2xcosπ/4-cos2xsinπ/4-2√2(1-cos2x)/2
=√2/2*sin2x-√2/2*cos2x+√2cos2x-√2
=√2/2*sin2x+√2/2*cos2x-√2
=sin2xcosπ/4+cos2xsinπ/4-√2
=sin(2x+π/4)-√2
T=2π/2=π
应是这样
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