Question about name-convention for secant and cosecant.

Question

Ok so if we take a right triangle and consider an angle $\alpha$ we get the following:

From here we can define the fundamental trigonometric functions sine and cosine where $\sin(\alpha)=\frac{\text{Opposite}}{\text{Hypotenuse}}$ and $\cos\alpha=\frac{\text{Adjacent}}{\text{Hypotenuse}}$

Now there are 2 new functions called secant and cosecant which are the multiplicative inverses of the sine and cosine functions.

What I don't understand is why $\csc(\alpha)=\frac{1}{\sin(\alpha)}$ and $ \sec(\alpha)=\frac{1}{\cos(\alpha)}$

Wouldn't it be easier to let the "co"-secant refer to the multiplicative inverse of the "co"-sine function?

Regards.

Answer 1

Yes, this would be nice if the "co"s matched up! However, it is indeed all geometry related. No calculus necessary, all you need is to look at the unit circle. Check out this webpage.

One thing to keep in mind is that a "secant line" is just a line that "cuts" through a figure. Secant has a root word in Latin (secare, I believe) which means "to cut". So, the secant should somehow involve "cutting" something, and the webpage above shows you the geometric idea behind the secant. It's the blue line that shows up.

Hopefully that makes it easier to forgive the co-confusion it invariably causes.

Answer 2

Defining "cosecant" to be the multiplicative inverse of "cosine" (and thus "secant" as the multiplicative inverse of "sine") might have been helpful from a "mnemonic" standpoint: using the "co" prefix to associate the two for memory purposes, but that's not how secant and cosecant are defined.

Note that there are many, many words that share a prefix but nonetheless have no relationship with one another: E.g., "butter", "butler", and "buttress" bear little in the way of similarity, save for the phenomenon that they all begin with the letters "but..."

One can remember to pair up $\,\csc x\,$ with $\dfrac 1{\sin x}$ by recalling that $\csc x$ is not the multiplicative inverse of $\cos x$, but rather of $\sin x$.