已知函数fx=sinxcosx-sin^2①求函数fx的最大值和最小值,并求出此时对应的想的值.②求函数fx在-pai/4,pai/4上的单调区间.

Question

已知函数fx=sinxcosx-sin^2
①求函数fx的最大值和最小值,并求出此时对应的想的值.
②求函数fx在【-pai/4,pai/4】上的单调区间.

Answer 1

fx=sinxcosx-sin²x
=1/2*2sinxcosx+(1-cos2x)/2
=1/2*sin2x+1/2-1/2cos2x
=1/2(sin2x-cos2x)+1/2
=√2/2sin(2x-π/4)+1/2
最大值=√2/2+1/2=(√2+1)/2
此时2x-π/4=π/2+2kπ,k∈Z
x=3π/8+kπ,k∈Z
最小值=-√2/2+1/2=(-√2+1)/2
此时2x-π/4=-π/2+2kπ,k∈Z
x=-π/8+kπ,k∈Z
(2)
x∈[-π/4,π/4]
2x∈[-π/2,π/2]
2x-π/4∈[-3π/4,π/4]
2x-π/4∈[-3π/4,-π/2]是sin(2x-π/4)的减区间
此时x∈[-π/4,-π/8]
∴综上
减区间是[-π/4,-π/8],增区间是[-π/8,π/4]