题目内容
已知函数f(x)=sin(x+
)+cos(x-
),x∈R
(Ⅰ)求f(x)的最小正周期和最小值;
(Ⅱ)已知cos(β-α)=
,cos(β+α)=-
.0<α<β≤
,求证:[f(β)]2-2=0.
| 7π |
| 4 |
| 3π |
| 4 |
(Ⅰ)求f(x)的最小正周期和最小值;
(Ⅱ)已知cos(β-α)=
| 4 |
| 5 |
| 4 |
| 5 |
| π |
| 2 |
(Ⅰ)f(x)=sin(x+
)+cos(x-
)=sin(x-
)+sin(x-
)=2sin(x-
)
∴T=2π,最小值为-2
(Ⅱ)∵cos(β-α)=cosβcosα+sinβsinα=
,cos(β+α)=cosβcosα-sinβsinα=-
,
两式相加得2cosβcosα=0,
∵0<α<β≤
,
∴β=
∴[f(β)]2-2=4sin2
-2=0
| 7π |
| 4 |
| 3π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| π |
| 4 |
∴T=2π,最小值为-2
(Ⅱ)∵cos(β-α)=cosβcosα+sinβsinα=
| 4 |
| 5 |
| 4 |
| 5 |
两式相加得2cosβcosα=0,
∵0<α<β≤
| π |
| 2 |
∴β=
| π |
| 2 |
∴[f(β)]2-2=4sin2
| π |
| 4 |
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