题目内容
已知α∈(
,
),β∈(0,
),且cos(
-α)=
,sin(
π+β)=-
求cos(α+β).
| π |
| 4 |
| 3π |
| 4 |
| π |
| 4 |
| π |
| 4 |
| 3 |
| 5 |
| 5 |
| 4 |
| 12 |
| 13 |
∵sin(
π+β)=-sin(
+β)=-
∴sin(
+β)=
∵β∈(0,
),
∴cos(
+β)=
∵α∈(
,
),cos(
-α)=
∴sin(
-α)=
cos(α+β)=cos[(
+β)-(
-α)]=
×
+
×
=
| 5 |
| 4 |
| π |
| 4 |
| 12 |
| 13 |
∴sin(
| π |
| 4 |
| 12 |
| 13 |
∵β∈(0,
| π |
| 4 |
∴cos(
| π |
| 4 |
| 5 |
| 13 |
∵α∈(
| π |
| 4 |
| 3π |
| 4 |
| π |
| 4 |
| 3 |
| 5 |
∴sin(
| π |
| 4 |
| 4 |
| 5 |
cos(α+β)=cos[(
| π |
| 4 |
| π |
| 4 |
| 5 |
| 13 |
| 3 |
| 5 |
| 12 |
| 13 |
| 4 |
| 5 |
| 63 |
| 65 |
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