题目内容
已知函数f(x)=
-
+
(k∈Z),求f(x)的值域.
| sin(kπ-x) |
| sinx |
| cosx |
| cos(kπ-x) |
| tan(kπ-x) |
| tanx |
当k=2n(k∈Z)时.f(x)=
-
+
=
-
+
=-1-1-1=-3.
当k=2n+1(n∈Z)时,f(x)=
-
+
=
-
+
=
-
+
=1+1-1=1
综上,当k∈Z时以z)的值域为{-3,1}.
| sin(2nπ-x) |
| sinx |
| cosx |
| cos(2nπ-x) |
| tan(2nπ-x) |
| tanx |
=
| sin(-x) |
| sinx |
| cosx |
| cos(-x) |
| tan(-x) |
| tanx |
当k=2n+1(n∈Z)时,f(x)=
| sin(2nπ+π-x) |
| sinx |
| cosx |
| cos(2nπ+π-x) |
| tan(2nπ+π-x) |
| tanx |
=
| sin(π-x) |
| sinx |
| cosx |
| cos(π-x) |
| tan(π-x) |
| tanx |
| sinx |
| sinx |
| cosx |
| -cosx |
| -tanx |
| tanx |
综上,当k∈Z时以z)的值域为{-3,1}.
练习册系列答案
相关题目