Question

函数y=x-2sinxcosx (-π/2

Answer 1

y=x-2sinxcosx

求导,则

y'=1-2(cos²x-sin²x)

=1-2cos2x

令y'=0,则1-2cos2x=0

解得x=kπ±(π/6)

由于x∈(-π/2,π/2),则

x=-π/6,x=π/6

由此可知

y'在(-π/2,-π/6)递增,在(-π/6,π/6)递减,在(π/6,π/2)递增

可见在x=-π/6处取得极大值,

此时

y=-π/6-2×(-1/2)×(√3/2)

=(√3/2)-π/6