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求函数y=sin(x+π/6)+cosx,(0≤x≤π)的最大值和最小值
人气:105 ℃ 时间:2019-10-20 19:29:37
解答
y=sin(x+π/6)+cosx
=sinxcos(π/6)+cosxsin(π/6)+cosx
=(√3/2)*sinx+(3/2)cosx
=√3*[(1/2)*sinx+(√3/2)cosx]
=√3*sin(x+π/3)
因为0≤x≤π,即π/3≤x+π/3≤4π/3
所以-√3/2≤sin(x+π/3)≤1
则当x+π/3=π/2,即x=π/6时,sin(x+π/3)=1,函数y有最大值√3;
当x+π/3=4π/3,即x=π时,sin(x+π/3)=-√3/2,函数y有最小值-3/2.
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