题目内容
化简sin(-2)+cos(2-π)?tan(2-4π)所得的结果是( )
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试题答案
B
相关题目
(1)化简:
(2)求值:sin
+cos
+tan(-
)+sin
.
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sin(2π-α)cos(π+α)cos(
| ||||
cos(π-α)sin(3π-α)sin(-π-α)sin(
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(2)求值:sin
| 25π |
| 6 |
| 23π |
| 3 |
| 25π |
| 4 |
| 4π |
| 3 |
(1)计算:
①cos0+5sin
-3sin
+10cosπ;
②cos
-tan
+
tan2
-sin
+cos2
+sin2
.
(2)化简:
.
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①cos0+5sin
| π |
| 2 |
| 3π |
| 2 |
②cos
| π |
| 3 |
| π |
| 4 |
| 3 |
| 4 |
| π |
| 6 |
| π |
| 6 |
| π |
| 4 |
| π |
| 3 |
(2)化简:
sin(2π-α)cos(3π+α)cos(
| ||
| sin(-π+α)sin(3π-α)cos(-α-π) |
(1)已知sin(α-3π)=2cos(α-4π),求
;
(2)化简
.
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| sin(π-α)+5cos(2π-α) | ||
2sin(
|
(2)化简
tan(π-α)cos(2π-α)sin(-α+
| ||
| cos(-α-π)sin(-π-α) |
(1)已知sin(α-3π)=2cos(α-4π),求
;
(2)化简
.
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| sin(π-α)+5cos(2π-α) | ||
2sin(
|
(2)化简
tan(π-α)cos(2π-α)sin(-α+
| ||
| cos(-α-π)sin(-π-α) |