题目
函数f(x)=sin^2x+sinxcosx的最小值和最大值分别为
提问时间:2020-12-06
答案
∵f(x)=sin^2x+sinxcosx
=(1-cos2x)/2+1/2sin2x
=1/2(sin2x-cos2x)+1/2
=√2/2sin(2x-π/4)+1/2
∴当sin(2x-π/4)=-1时
f(x)min=(1-√2)/2
当sin(2x-π/4)=1时
f(x)max=(1+√2)/2
=(1-cos2x)/2+1/2sin2x
=1/2(sin2x-cos2x)+1/2
=√2/2sin(2x-π/4)+1/2
∴当sin(2x-π/4)=-1时
f(x)min=(1-√2)/2
当sin(2x-π/4)=1时
f(x)max=(1+√2)/2
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