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已知函数f(x)=sin(2x+派/4)+sin(2x-派/4)(1)化简并求出最小正周期.
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已知函数f(x)=sin(2x+派/4)+sin(2x-派/4) (1)化简并求出最小正周期.
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答案和解析
f(x)=sin(2x+π/4)+sin(2x-π/4)
=2sin(2x+π/4+2x-π/4)/2cos(2x+π/4-2x+π/4)/2
=2sin2xcosπ/4
=√2sin2x
所以最小正周期T=2π/2=π
=2sin(2x+π/4+2x-π/4)/2cos(2x+π/4-2x+π/4)/2
=2sin2xcosπ/4
=√2sin2x
所以最小正周期T=2π/2=π
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