Question

已知函数y=sin^2x+2sinxcosx+3cos^2*(3兀+x)

Answer 1

y=sin^2x+2sinxcosx+3cos^2*(3兀+x)
=sin^2x+2sinxcosx+3cos^2x
=2cos²x+1+sin2x
=cos2x+2+sin2x
=√2sin(2x+π/4)+2
最小正周期
为
2π/2=π
单增
区间:
2x+π/4∈[2kπ-π/2,2kπ+π/2]
x∈[kπ-3π/8,kπ+π/8]
单调减区间为
[kπ+π/8,kπ+5π/8]
x∈[-π/2,0]
2x+π/4∈[-3π/4,π/4]
所以
sin(2x+π/4)的范围为[-1,√2/2]
所以
值域
为[-√2+2,3]
最大值为3
最小值为
2-√2

Answer 2

1、
y=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1
=1+sin2x+cos2x+1
=√2sin(2x+π/4)+2
所以t=2π/2=π
2、
sin的系数为正
所以y的单调性和sin(2x+π/4)相同
sinx的增区间是(2kπ-π/2,2kπ+π/2)
所以2kπ-π/2<2x+π/4<2kπ+π/2
2kπ-3π/4<2x<2kπ+π/4
kπ-3π/8

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