Help with trigonometry? My question says to find the range of the following functions:
$$y = \csc x$$
$$y = \sec x$$
$$y = \cot x$$
Only I don't know what they mean by range and how to find the answer.
 A: The range is all the $y$-values the functions can possibly take.  To get a feel for the range you can graph the function or plug in a few points.  But that doesn't prove anything.
To find the range of $\csc(x) = \frac{1}{\sin(x)}$, remember that


*

*$\sin(x)$ can only lie between $-1$ and $1$, but also includes $0$.

*If you consider only positive values of $\sin(x)$, $\frac{1}{\sin(x)}$ gets smaller the bigger $\sin(x)$ gets.  It can't go below $1$ since $\sin(x)$ can't be greater than $1$.

*When $\sin(x)$ approaches $0$ from the positive direction, $\frac{1}{\sin(x)}$ becomes large and positive.  In fact, since $\frac{1}{x}$ is continuous at nonzero values of $x$, $\frac{1}{\sin(x)}$ takes on all values between $1$ and $\infty$.

*For the negative values of $\sin(x)$ you can do the same thing.  $\sin(x)$ cannot be less than $-1$, so (for negative values), $\frac{1}{\sin(x)}$ can't go above $-1$.  Since $\frac{1}{\sin(x)}$ goes to $-\infty$ as $\sin(x)$ approaches $0$ from the negative direction, it takes on all values between $-\infty$ and $-1$.

*So the range of $\csc(x)$ is $(-\infty, -1] \cup [1, \infty)$.


Try doing the rest yourself!  Hope that helped
