Question

高二数学，急求大神帮我求y＝sin2x+cos2x的最小正周期，最大最小值，和单调区间(过程要详细)

高二数学，急求大神帮我求y＝sin2x+cos2x的最小正周期，最大最小值，和单调区间(过程要详细)谢谢了

Answer 1

解：
y＝sin2x+cos2x
=√2(sin2x*√2/2+cos2x*√2/2)
=√2sin(2x+π/4)
最小正周期T=2π/w=2π/2=π

∵-1≤sin(2x+π/4)≤1
∴-√2≤√2sin(2x+π/4)≤√2
最大值为√2
最小值为-√2

令-π/2+2kπ≤2x+π/4≤π/2+2kπ
-3π/4+2kπ≤2x≤π/4+2kπ
-3π/8+kπ≤x≤π/8+2kπ
所以单调递增区间为｛x l -3π/8+kπ≤x≤π/8+2kπ｝(k∈Z)
令π/2+2kπ≤2x+π/4≤3π/2+2kπ
π/4+2kπ≤2x≤5π/4+2kπ
π/8+kπ≤x≤5π/8+2kπ
所以单调递减区间为｛x l π/8+kπ≤x≤5π/8+2kπ｝(k∈Z)

Answer 2

将y＝sin2x+cos2x化成一角一函数的形式y=√2sin（2x+π/4）=√2sin[2（x+π/8）]
A=√2，ω=2，ψ=π/4
最小正周期T=2π/ω=π
最大值为A=√2
最小值为-A=-√2
2x+π/4=-π/2+2kπ，x=-3π/8+kπ
2x+π/4=π/2+2kπ，x=π/8+kπ
2x+π/4=3π/2+2kπ，x=5π/8+kπ
增区间(-3π/8+kπ,π/8+kπ)
减区间(π/8+kπ,5π/8+kπ)
k为整数

Answer 3

y=sin2x+cos2x
=sin2x+sin(π/2-2x)
=2sinπ/4cos(2x-π/4)
所以最小正周期=2π/2=π
最大值=2sinπ/4=√2
最小值=-2sinπ/4=-√2
单调递增区间: 2kπ-π/2<=2x-π/4<=2kπ+π/2, kπ-π/8<=x<=kπ+3π/8
单调递减区间: 2kπ+π/2<=2x-π/4<=2kπ+3π/2,kπ+3π/8<=x<=kπ+7π/8

Answer 4

Y=Sin2x+cos2x=根号2（根号2/2sin2x+根号2/2cos2x)=根号2sin(45度+2x )
所以最小正周期是π
因为sin最大值是1，最小值是-1
所以此题最大值是根号2，最小值是-根号2

Answer 5

π 最大值根号2，最小值﹣根号2