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化简1+sinx-cosx-sin2x/1+sinx-cosx,并求出其最大值.
人气:417 ℃ 时间:2020-04-19 04:51:15
解答
(1+sinx-cosx-sin2x)/(1+sinx-cosx)
=(1+sinx-cosx-sin2x)/(1+sinx-cosx);
=(sin^2x+cos^2x+sinx-cosx-2sinxcosx)/(1+sinx-cosx);
=(sin^2x-2sinxcosx+cos^2x+sinx-cosx)/(1+sinx-cosx);
=[(sinx-cosx)^2+(sinx-cosx)]/(1+sinx-cosx);
=(sinx-cosx)*(sinx-cosx+1)/(sinx-cosx+1)
=sinx-cosx
=√2sin(x-π/4)
-√2<=√2sin(x-π/4)<=√2
所以(1+sinx-cosx-sin2x)/(1+sinx-cosx)的最大值为:√2
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