题目内容
(1)已知cos(x+
)=
,求cos(
-x)+cos2(
-x)的值;
(2)计算:sin
+cos2
cosπ+3tan2
+cos
-sin
.
| π |
| 6 |
| 1 |
| 4 |
| 5π |
| 6 |
| π |
| 3 |
(2)计算:sin
| π |
| 6 |
| π |
| 4 |
| π |
| 6 |
| π |
| 3 |
| π |
| 2 |
(1)∵cos(x+
)=
,∴cos(
-x)=cos[π-(x+
)]=-cos(x+
)=-
,
cos(
-x)=cos[
-(x+
)]=sin(x+
),
∴cos2(
-x)=sin2(x+
)=1-cos2(x+
)=1-
=
.
∴cos(
-x)+cos2(
-x)=-
+
=
.
(2)sin
+cos2
cosπ+3tan2
+cos
-sin
=
+
×(-1)+3×
+
-1=
.
| π |
| 6 |
| 1 |
| 4 |
| 5π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| 1 |
| 4 |
cos(
| π |
| 3 |
| π |
| 2 |
| π |
| 6 |
| π |
| 6 |
∴cos2(
| π |
| 3 |
| π |
| 6 |
| π |
| 6 |
| 1 |
| 16 |
| 15 |
| 16 |
∴cos(
| 5π |
| 6 |
| π |
| 3 |
| 1 |
| 4 |
| 15 |
| 16 |
| 11 |
| 16 |
(2)sin
| π |
| 6 |
| π |
| 4 |
| π |
| 6 |
| π |
| 3 |
| π |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 2 |
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