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求y=sin(2x+4分之π)+cos(2x-4分之π)的最小正周期
人气:487 ℃ 时间:2020-04-06 01:49:52
解答
y=sin(2x+4分之π)+cos(2x-4分之π)
=sin(2x+π/4)+cos[-π/2+(π/4+2x)]
=sin(2x+π/4)+sin(2x+π/4)
=2sin(2x+π/4)
∴ 求y=sin(2x+4分之π)+cos(2x-4分之π)的最小正周期是T=2π/2=π
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