题目内容
已知f(x)=2sin(x+
),x∈R.
(1)求f(
)的值;
(2)设α,β∈∈[0,
],f(α)=2,f(β)=
,求f(α+β)的值.
| π |
| 6 |
(1)求f(
| π |
| 6 |
(2)设α,β∈∈[0,
| π |
| 3 |
| 8 |
| 5 |
考点:两角和与差的正弦函数
专题:三角函数的求值
分析:(1)直接根据函数f(x)的解析式求得f(
)的值.
(2)根据f(α)=2,求得sin(x+
)的值,可得α的值.根据 f(β)=
,求得sin(β+
)的值,可得cos(β+
)的值.再根据f(α+β)=2sin(α+β+
)=2cosβ=2cos[(β+
)-
],利用两角和差的余弦公式求得f(α+β)的值.
| π |
| 6 |
(2)根据f(α)=2,求得sin(x+
| π |
| 6 |
| 8 |
| 5 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
解答:
解:(1)f(
)=2sin(
+
)=2sin
=
.
(2)∵α,β∈[0,
],f(α)=2sin(α+
)=2,∴sin(α+
)=1,
∴α+
=
,α=
.
∵f(β)=2sin(β+
)=
,∴sin(β+
)=
,
∵β+
∈[
,
],∴cos(β+
)=
.
∴f(α+β)=2sin(α+β+
)=2sin(
+β)=2cosβ
=2cos[(β+
)-
]=2cos(β+
)cos
+2sin(β+
)sin
=2×
×
+2×
×
=
.
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 3 |
| 3 |
(2)∵α,β∈[0,
| π |
| 3 |
| π |
| 6 |
| π |
| 6 |
∴α+
| π |
| 6 |
| π |
| 2 |
| π |
| 3 |
∵f(β)=2sin(β+
| π |
| 6 |
| 8 |
| 5 |
| π |
| 6 |
| 4 |
| 5 |
∵β+
| π |
| 6 |
| π |
| 6 |
| π |
| 2 |
| π |
| 6 |
| 3 |
| 5 |
∴f(α+β)=2sin(α+β+
| π |
| 6 |
| π |
| 2 |
=2cos[(β+
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
=2×
| 3 |
| 5 |
| ||
| 2 |
| 4 |
| 5 |
| 1 |
| 2 |
3
| ||
| 5 |
点评:本题主要考查同角三角函数的基本关系、两角和差的正弦公式的应用,属于中档题.
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