已知,函数y=(sinx+cosx)²+2cos²x (1)求它的递减区间 (2)求它的最大值与最小值
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解 y=(sinx+cosx)²+2cos²x
=2+sin2x +cos2x
=√2sin(2x+π/4)+2
(1)由 2kπ+π/2≤2x+π/4≤2kπ+3π/2
得 kπ+π/8 ≤ x≤kπ+5π/8
故递减区间为[kπ+π/8 ,kπ+5π/8]
(2)当2x+π/4=2kπ+π/2 即 x=kπ+π/8 时sin(2x+π/4)=1
y有最大值√2+2;
当2x+π/4=2kπ+3π/2 即 x=kπ+5π/8 时sin(2x+π/4)=-1
y有最小值 2-√2
=2+sin2x +cos2x
=√2sin(2x+π/4)+2
(1)由 2kπ+π/2≤2x+π/4≤2kπ+3π/2
得 kπ+π/8 ≤ x≤kπ+5π/8
故递减区间为[kπ+π/8 ,kπ+5π/8]
(2)当2x+π/4=2kπ+π/2 即 x=kπ+π/8 时sin(2x+π/4)=1
y有最大值√2+2;
当2x+π/4=2kπ+3π/2 即 x=kπ+5π/8 时sin(2x+π/4)=-1
y有最小值 2-√2
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