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求函数y=cosx(cosx+sinx)的值域
如题~
人气:346 ℃ 时间:2019-12-18 07:42:38
解答
y=cosx(cosx+sinx)
=cos²x+sinxcosx
=(cos2x+1)/2+1/2·sin2x
=1/2·(sin2x+cos2x)+1/2
=1/2·√2(√2/2·sin2x+√2/2·cos2x)+1/2
=1/2·√2(cosπ/4·sin2x+sinπ/4·cos2x)+1/2
=1/2·√2sin(2x+π/4)+1/2
=√2/2·sin(2x+π/4)+1/2
∵sin(2x+π/4)∈【-1,1】
∴1/2-√2/2≤√2/2·sin(2x+π/4)+1/2≤1/2+√2/2
故值域为:【1/2-√2/2,1/2+√2/2】=cos²x+sinxcosx =(cos2x+1)/2+1/2·sin2x这步是怎么变的?根据cos2x=2cos²x-1得cos²x=(co2x+1)/2还有sin2x=2sinxcosx
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