(2005•闸北区一模)已知函数f(x)=sin(x+π6)−cos(x+π3)+cosx,
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(2005•闸北区一模)已知函数f(x)=sin(x+
)−cos(x+
)+cosx
π |
6 |
π |
3 |
(Ⅰ)f(x)=sin(x+
π
6)−cos(x+
π
3)+cosx
=
3
2sinx+
1
2cosx-(
1
2cosx-
3
2sinx)+cosx
=
3sinx+cosx
=2sin(x+
π
6),
∵ω=1,∴T=2π,
令2kπ+
π
2≤x+
π
6≤2kπ+
3π
2,解得:2kπ+
π
3≤x≤2kπ+
4π
3,
则函数的单调递减区间:[2kπ+
π
3,2kπ+
4π
3](k∈Z);
(Ⅱ)x∈[−
π
2,
π
2]⇒x+
π
6∈[−
π
3,
2π
3]⇒M=f(
π
3)=2,m=f(−
π
2)=−
π
6)−cos(x+
π
3)+cosx
=
3
2sinx+
1
2cosx-(
1
2cosx-
3
2sinx)+cosx
=
3sinx+cosx
=2sin(x+
π
6),
∵ω=1,∴T=2π,
令2kπ+
π
2≤x+
π
6≤2kπ+
3π
2,解得:2kπ+
π
3≤x≤2kπ+
4π
3,
则函数的单调递减区间:[2kπ+
π
3,2kπ+
4π
3](k∈Z);
(Ⅱ)x∈[−
π
2,
π
2]⇒x+
π
6∈[−
π
3,
2π
3]⇒M=f(
π
3)=2,m=f(−
π
2)=−
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