Question

高分速度急！急！跪求！已知函数y=(sinx+cosx)的平方+2cos的平方x求最小正周期！递减区间！最大最小值！

Answer 1

y=sin²x+cos²x+2sinxcosx+2(1+cos2x)/2
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
所以最大值=√2+2
最小值=√2/2
递减则2kπ+π/2<2x+π/4<2kπ+3π/2
kπ+π/8<x<kπ+5π/8
所以减区间(kπ+π/8,kπ+5π/8)

Answer 2

y=sin²x+cos²x+2sinxcosx+2cos²x
=√2sin(2x+π/4)+2

所以T=2π/2=π

sin递减则2kπ+π/2<2x+π/4<2kπ+3π/2
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 3

y=1+sin2x+cos2x-1=根号2sin(2x+pi/4)
T=pi
递减区间 2k*pi+pi/2<=2x+pi/4<=2k*pi+3Pi/2 k*pi+pi/8<=x<=k*pi5pi/8

最大值 根号2
最小值 -根号2

Answer 4

y=(sinx+cosx)^2+2(cosx)^2
=1+sin(2x)+1+cos(2x)
=2+√2sin(2x+π/4)，周期是π
递增区间kπ-3π/8≤x≤kπ+π/8 (k∈Z)
单减区间kπ+π/8≤x≤kπ+5π/8 (k∈Z)
sin(2x+π/4)=±1，ymax=2+√2，ymin=2-√2

Answer 5

y=1+sin2x+2cos^2 (x)
=sin2x+cos2x+2
=根号2*sin(2x+π/4）+2
周期T=2π/2=π
递减区间π/2+2kπ<=(2x+π/4）<=3π/2+2kπ，
得π/8+kπ<=x<=5π/8+kπ
最大值=根号2*1+2，最小值=根号2*（-1）+2

Answer 6

化简呗。结果是1+sin(2x)+1+cos(2x)=2+（根号（2））*sin(2x+pi/4)

就化成了求sin(2x+pi/4)的问题了 这就简单了呗