# 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.

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bit.ly/a8GLzI –  user996522 Mar 17 '12 at 15:37
–  Jeremy Carlos Mar 17 '12 at 15:40
I hope you will not think I am being rude or sarcastic if I suggest that you look to see if your book has an index, and if so if the index has an entry for "range". –  MJD Mar 18 '12 at 1:49

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

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you are good at math –  Jeremy Carlos Mar 20 '12 at 21:06