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求f(x)=2cos^2x+sin2x的最值小
人气:360 ℃ 时间:2020-05-12 17:41:59
解答
f(x)=2cos^2x+sin2x
=1+cos2x+sin2x
=sin2x+cos2x+1
=√2*(√2/2*sin2x+√2/2*cos2x)+1
=√2*(sin2xcosπ/4+cos2xsinπ/4)+1
=√2*sin(2x+π/4)+1
-1
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