函数y=cos(2x-3π/4)-2√2(sinx)^2的最小正周期为_____.y=cos(2x-3π/4)-2√2(sinx)^2=cos(2x-3π/4)+√2[1-2(sinx)^2]-√2=-√2/2*cos2x+√2/2*sin2x+√2cos2x- √2=√2/2*cos2x+√2/2*sin2x-√2=cos(2x-pi/4)-√2=-√2/2*cos2x+√2/2*sin2x+
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函数y=cos(2x-3π/4)-2√2(sinx)^2的最小正周期为_____.y=cos(2x-3π/4)-2√2(sinx)^2=cos(2x-3π/4)+√2[1-2(sinx)^2]-√2=
函数y=cos(2x-3π/4)-2√2(sinx)^2的最小正周期为_____.y=cos(2x-3π/4)-2√2(sinx)^2=cos(2x-3π/4)+√2[1-2(sinx)^2]-√2=-√2/2*cos2x+√2/2*sin2x+√2cos2x- √2=√2/2*cos2x+√2/2*sin2x-√2=cos(2x-pi/4)-√2=-√2/2*cos2x+√2/2*sin2x+
函数y=cos(2x-3π/4)-2√2(sinx)^2的最小正周期为_____.
y=cos(2x-3π/4)-2√2(sinx)^2
=cos(2x-3π/4)+√2[1-2(sinx)^2]-√2
=-√2/2*cos2x+√2/2*sin2x+√2cos2x- √2
=√2/2*cos2x+√2/2*sin2x-√2
=cos(2x-pi/4)-√2
=-√2/2*cos2x+√2/2*sin2x+√2cos2x- √2
=√2/2*cos2x+√2/2*sin2x-√2
之中,+√2cos2x哪儿去了?
函数y=cos(2x-3π/4)-2√2(sinx)^2的最小正周期为_____.y=cos(2x-3π/4)-2√2(sinx)^2=cos(2x-3π/4)+√2[1-2(sinx)^2]-√2=-√2/2*cos2x+√2/2*sin2x+√2cos2x- √2=√2/2*cos2x+√2/2*sin2x-√2=cos(2x-pi/4)-√2=-√2/2*cos2x+√2/2*sin2x+
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