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求函数y=2cos(x+π/4)cos(x-π/4)+更3sin2x的值域和最小正周期.3Q
人气:453 ℃ 时间:2019-09-29 01:22:52
解答
y=2cos(x+π/4)*cos(x-π/4)+(√3)sin2x =2sin(π/4-x)*cos(π/4-x)+(√3)sin2x =sin(π/2-2x)+(√3)sin2x =cos2x+(√3)sin2x =2[cosπ/3cos2x+sinπ/3sin2x] =2cos(2x-π/3) 所以,值域[-2,2],周期为π,2kπ
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