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已知函数y=1-2sin^2x+2sinxcosx,求:
(1)该函数的最小正周期;
(2)该函数的值域.
人气:200 ℃ 时间:2019-10-11 05:01:20
解答
y=cos2x+sin2x
=√2(√2/2*sin2x+√2/2cos2x)
=√2(sin2xcosπ/4+cos2xsinπ/4)
=√2sin(2x+π/4)
所以T=2π/2=π
值域是[-√2,√2]
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