问题
解答题
已知△ABC的面积S满足3≤S≤3
(Ⅰ)求θ的取值范围; (Ⅱ)求函数f(θ)=sin2θ+2sinθcosθ+3cos2θ的最大值. |
答案
(I)由题意知
•AB
=|BC
||AB
|cosθ=6.BC
S=
|1 2
| |AB
|sin(π-θ)=BC
|1 2
| |AB
|sinθBC
=
|1 2
| |AB
|cosθtanθBC
=
×6tanθ=3tanθ.1 2
∵3≤S≤3
,3
∴3≤3tanθ≤3
,∴1≤tanθ≤3
.3
又∵θ∈[0,π],∴
≤θ≤π 4
.π 3
(II)∵f(θ)=sin2θ+2sinθcosθ+3cos2θ
=1+sin2θ+2cos2θ
=2+sin2θ+cos2θ=2+
sin(2θ+2
).π 4
,∴(2θ+∵θ∈[
,π 4
]π 3
)∈[π 4
,3π 4
].11π 12
∵y=sinx在[
,π]上单调递减,π 2
∴当2θ+
=π 4
,即θ=3π 4
时,sin(2θ+π 4
)取得最大值π 4
,2 2
∴f(θ)的最大值为2+
×2
=3.2 2