Question

求函数y＝2sinxcosx＋cos2x的定义域和单调区间

Answer 1

y=2sinxcosx+cos2x
=sin2x+cos2x
=√2(√2/2sin2x+√2/2cos2x)
=√2sin(2x+π/4)
所以可得定义域为：R
单调递增区间为：[kπ-3π/8,kπ+π/8]
单调递减区间为：[kπ+π/8,kπ+5π/8]

Answer 2

y＝2sinxcosx＋cos2x
=sin2x+cos2x
=√2(sin2x * √2/2 +cos2x * √2/2)
=√2(sin2xcosπ/4+cos2xsinπ/4)
=√2sin(2x+π/4)
所以定义域为一切实数
增区间
2kπ+π/2>=2x+π/4>=2kπ-π/2
2kπ+π/4>=2x>=2kπ-3π/4
kπ+π/8>=x>=kπ-3π/4
减区间
2kπ+π3/2>=2x+π/4>=2kπ+π/2
2kπ+5π/4>=2x>=2kπ+π/4
kπ+5π/8>=x>=kπ+π/8