题目内容
在△ABC中,A,B,C为三个内角,f(x)=4cosxsin2(
+
)+
cos2x-2cosx.
(1)若f(B)=2,求角B;
(2)若f(B)-m>2有解,求实数m的取值范围;
(3)求f(
)+f(
)+f(
)+…+f(
)的值.
| π |
| 4 |
| x |
| 2 |
| 3 |
(1)若f(B)=2,求角B;
(2)若f(B)-m>2有解,求实数m的取值范围;
(3)求f(
| π |
| 4 |
| 2π |
| 4 |
| 3π |
| 4 |
| 2003π |
| 4 |
(1)∵sin2(
+
)=
=
,
∴f(x)=4cosx×
+
cos2x-2cosx
=2cosx+sin2x+
cos2x-2cosx
=2sin(2x+
).
∵f(B)=2,∴2sin(2B+
)=2,∴sin(2B+
)=1.
∵0<B<π,∴
<2B+
<2π+
,
∴2B+
=
,解得B=
.
(2)由(1)可知:f(B)∈[-2,2],
∵f(B)-m>2有解,∴2+m<[f(B)]max,∴2+m<2,解得m<0.
∴m的取值范围是(-∞,0).
(3)∵f(x)的周期是π,且f(
)+f(
)+f(
)+f(π)=2[sin(
+
)+sin(π+
)+sin(
+
)+sin(2π+
)]
=2[cos
-sin
-cos
+sin
]=0.
∴f(
)+f(
)+f(
)+…+f(
)
=500×4×0+f(
)+f(
)+f(
)=f(
)+f(
)+f(
)
=2×(-sin
)=-
.
| π |
| 4 |
| x |
| 2 |
1-cos(
| ||
| 2 |
| 1+sinx |
| 2 |
∴f(x)=4cosx×
| 1+sinx |
| 2 |
| 3 |
=2cosx+sin2x+
| 3 |
=2sin(2x+
| π |
| 3 |
∵f(B)=2,∴2sin(2B+
| π |
| 3 |
| π |
| 3 |
∵0<B<π,∴
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
∴2B+
| π |
| 3 |
| π |
| 2 |
| π |
| 12 |
(2)由(1)可知:f(B)∈[-2,2],
∵f(B)-m>2有解,∴2+m<[f(B)]max,∴2+m<2,解得m<0.
∴m的取值范围是(-∞,0).
(3)∵f(x)的周期是π,且f(
| π |
| 4 |
| 2π |
| 4 |
| 3π |
| 4 |
| π |
| 2 |
| π |
| 3 |
| π |
| 3 |
| 3π |
| 2 |
| π |
| 3 |
| π |
| 3 |
=2[cos
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
| π |
| 3 |
∴f(
| π |
| 4 |
| 2π |
| 4 |
| 3π |
| 4 |
| 2003π |
| 4 |
=500×4×0+f(
| 2001π |
| 4 |
| 2002π |
| 4 |
| 2003π |
| 4 |
| π |
| 4 |
| 2π |
| 4 |
| 3π |
| 4 |
=2×(-sin
| π |
| 3 |
| 3 |
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