题目内容
已知f(x)=[sin(x+
)+
cos(x+
)]•cos(x+
).若θ∈[0,π]且f(x)为偶函数,求θ的值.
| θ |
| 2 |
| 3 |
| θ |
| 2 |
| θ |
| 2 |
分析:通过多项式展开,利用二倍角已以及两角和的正弦函数,化简函数的表达式,通过偶函数的定义求出θ的值.
解答:解:f(x)=[sin(x+
)+
cos(x+
)]•cos(x+
)
=sin(x+
)•cos(x+
)+
cos2(x+
)
=
sin(2x+θ)+
[1+cos(2x+θ)]
=sin(2x+θ+
)+
因为f(x)为偶函数,所以f(-x)=f(x),
即sin(-2x+θ+
)=sin(2x+θ+
),得sin2x•cos(θ+
)=0,
所以cos(θ+
)=0.又θ∈[0,π],所以θ=
.
| θ |
| 2 |
| 3 |
| θ |
| 2 |
| θ |
| 2 |
=sin(x+
| θ |
| 2 |
| θ |
| 2 |
| 3 |
| θ |
| 2 |
=
| 1 |
| 2 |
| ||
| 2 |
=sin(2x+θ+
| π |
| 3 |
| ||
| 2 |
因为f(x)为偶函数,所以f(-x)=f(x),
即sin(-2x+θ+
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
所以cos(θ+
| π |
| 3 |
| π |
| 6 |
点评:本题考查三角函数的化简求值,二倍角以及两角和的正弦函数的应用,考查计算能力.
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