已知函数y=4cos2x(sin2x
已知函数y=4cos2x(sin2x-cos2x),求函数的值域需详细过程。
y=4cos2x(sin2x-cos2x) =4cos2xsin2x-4(cos2x)^2 =2sin4x-2(1+cos4x) =2sin4x-2cos4x-2 =2(√2)sin(4x-π/4)-2 -2(√2)-2≤y≤2(√2)-2 值域[-2-2√2,-2+2√2]
y=4cos2x(sin2x-cos2x) =4sin2xcos2x-4(cos2x)^2 =2sin4x-2(1+cos4x) =2(sin4x-cos2x)-2 =2√2sin(4x-45°)-2, 其值域是[-2-2√2,2√2-2].
原式=4cos(2x)sin(2x)-4[cos(2x)]^2=2sin(4x)-2[cos(4x)+1] =2[sin(4x)-cos(4x)]-2 =2*根号2*sin(4x-pi/4)-2 值域是 [2*根号2-2 2*根号2+2]
答:求函数y=sin2x+cos2x 单调递减 解:由化一公式y=sin2x+cos2x=√2sin(2x+π/4) 因为 y=sinx的单调递减区间[2kpi+p...详情>>
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