2sin^4x+2cos^4x-cos^2x-3的最小正周期.
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2sin^4x+2cos^4x-cos^2x-3的最小正周期.2sin^4x+2cos^4x-cos^2x-3的最小正周期.2sin^4x+2cos^4x-cos^2x-3的最小正周期.首先将f(x)
2sin^4x+2cos^4x-cos^2x-3的最小正周期.
2sin^4x+2cos^4x-cos^2x-3的最小正周期.
2sin^4x+2cos^4x-cos^2x-3的最小正周期.
首先将f(x)=2sin^4 x+2cos^4 x+cos^2 x-3 化简,得到
f(x)=cos^2(2x)+cos(2x)/2-3/2
=cos(4x)/2+cos(2x)/2-1,
函数f(x)的最小正周期显然等于cos(2x)的周期π.
2sin^4x+2cos^4x-cos^2x-3
=2[(sin^4x+cos^4x-2sin^2xcos^2x)+2sin^2xcos^2x]-cos^2x-3
=2[(sin^2x+cos^2x)^2-2sin^2xcos^2x]-cos^2x-3
=2(1-2sin^2xcos^2)-cos^2x-3
=2-4sin^2xcos^2x-cos^2x-3
=2-(2sinxcosx)^2-cos^2x-3
=2-sin^2(2x)-1/2[(1+cos2x)]-3
=-sin^2(2x)-1/2cos2x-3/2
………………呃,不会了。
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