题目内容
| π |
| 2 |
(1)写出f(x)的解析式及x0的值;
(2)若锐角θ满足cosθ=
| 1 |
| 3 |
分析:(1)由函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<
)的图象可求得A,ω,及φ的值,从而可求得f(x)的解析式及x0的值;
(2)依题意可求得sinθ的值,而f(2θ)=2sin(θ+
),利用两角和的正弦即可求得答案.
| π |
| 2 |
(2)依题意可求得sinθ的值,而f(2θ)=2sin(θ+
| π |
| 6 |
解答:解:(1)由函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<
)的图象可知A=2,
=2π,
∴
=4π,
∴ω=
;
又f(0)=1,
∴2sinφ=1,而|φ|<
,
∴φ=
.
∴f(x)=2sin(
x+
);
又
x0+
=
,
∴x0=
.
(2)∵cosθ=
,θ为锐角,
∴sinθ=
,
∴f(2θ)=2sin(
×2θ+
)
=2sin(θ+
)
=2(sinθcos
+cosθsin
)
=2(
×
+
×
)
=
.
| π |
| 2 |
| T |
| 2 |
∴
| 2π |
| ω |
∴ω=
| 1 |
| 2 |
又f(0)=1,
∴2sinφ=1,而|φ|<
| π |
| 2 |
∴φ=
| π |
| 6 |
∴f(x)=2sin(
| 1 |
| 2 |
| π |
| 6 |
又
| 1 |
| 2 |
| π |
| 6 |
| π |
| 2 |
∴x0=
| 2π |
| 3 |
(2)∵cosθ=
| 1 |
| 3 |
∴sinθ=
2
| ||
| 3 |
∴f(2θ)=2sin(
| 1 |
| 2 |
| π |
| 6 |
=2sin(θ+
| π |
| 6 |
=2(sinθcos
| π |
| 6 |
| π |
| 6 |
=2(
2
| ||
| 3 |
| ||
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
=
2
| ||
| 3 |
点评:本题考查由y=Asin(ωx+φ)的部分图象确定其解析式,考查三角函数的恒等变换及化简求值,属于中档题.
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