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y=sin^2x+sinxcosx+2cos^2x的值域
人气:440 ℃ 时间:2020-09-12 15:41:13
解答
y=sin^2x+sinxcosx+2cos^2x
=sinxcosx+cos^2x+1
=(1/2)(sin2x+cos2x)+3/2
=(√2/2)sin(2x+45°)+3/2
∵-1≤sin(2x+45°)≤1
∴y∈[3/2-√2/2,3/2+√2/2]
即为所求
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