Question

已知函数y=(sinx+cosx)平方+2cos平方x.求它的递减区间和最值。

已知函数y=(sinx+cosx)平方+2cos平方x.求它的递减区间和最值。

Answer 1

y=sin²x+cos²x+2sinxcosx+2cos²x-1+1
=1+sin2x+cos2x+1
=√2sin(2x+π/4)+2
-1<=sin(2x+π/4)<=1
所以显然最值是±√2+2

sinx的减区间是(2kπ+π/2,2kπ+3π/2)
所以2kπ+π/2<2x+π/4<2kπ+3π/2
2kπ+π/4<2x<2kπ+5π/4
kπ+π/8<x<kπ+5π/8

所以减区间(kπ+π/8,kπ+5π/8)
最大√2+2,最小-√2+2

Answer 2

y=sin²x+cos²x+2sinxcosx+2cos²x-1+1
=1+sin2x+cos2x+1
=√2sin(2x+π/4)+2
-1<=sin(2x+π/4)<=1
所以显然最值是±√2+2
sinx的减区间是(2kπ+π/2,2kπ+3π/2)
所以2kπ+π/2<2x+π/4<2kπ+3π/2
2kπ+π/4<2x<2kπ+5π/4
kπ+π/8<x<kπ+5π/8
所以减区间(kπ+π/8,kπ+5π/8)
最大√2+2,最小-√2+2