# Thread: Deriving an inverse of a function which has a trigonometric function inside of it

1. ## Deriving an inverse of a function which has a trigonometric function inside of it

This problem is taken straight from a free textbook of Mooculus. The problem itself has nothing to do with calculus as it's part of introductory review section of the book. I hope it goes into right sub-forum. If not, I kindly ask one of the site admins to more it to right section, if possible.

This is not the original problem and the question presented in the book (page 17, exercise 5 to be exact) is easily solvable by just looking at graphs, so my interest here is purely out of curiosity and for learning.

Now that the preliminaries are out of the way, onwards to the actual problem.

Let us have a function ie. f(x) = 18*sin((pi*x)/7)+20 and let us also restrict it to [3.5,10.5] in order to make it inversible. With my feeble algebraic skills I can come up with ie. f^1-(x) = 7 * arcsin((t - 20)/18)/pi which is blatantly incorrect. From another source I got the right answer which is ie. 7 * (-arcsin((t - 20)/18) + pi)/pi. What I do not understand is how to get to this. I know this has everything to do with the fact that arcsin is only defined in [-pi/2, pi/2], but I can't quite work this out on my own. Could someone explain the process in as simple steps as possible? After two days of occasional poking at this I feel like being on the verge of "finally getting it", yet I can't just quite grasp what on earth I'm actually pondering about.

2. ## Re: Deriving an inverse of a function which has a trigonometric function inside of it

Hey tlichy.

Can you show us stepe by step your path to finding the inverse?

In terms of arc-sin being defined on [-pi/2,pi/2] the reason is because outside of these values you get ambiguous values. Basically a function has an inverse if both the horizontal and vertical line tests are passed. Normal functions must only pass the vertical test, but invertible functions must pass the horizontal line test. The line test states that if you draw a line at any point of the function, then you only get one value that intersects it.

We define this interval because it is natural to do so. The arc-sin formula for this branch makes sense (you need to understand taylor series to understand this statement). We can only have one branch which is pi units in length and the branch [-pi/2,pi/2] is the most appropriate.

If your range is outside this, you must adjust for the arc-sin function to make sense. The arc-sin takes value ths from -1 to +1 and maps them from -pi/2 to pi/2. If your angles are different then you need to adjust them by adding the appropriate multiple of 2*k*pi where k is an integer. The reason is that sin(x + 2*k*pi) = sin(x) for any integer k.

3. ## Re: Deriving an inverse of a function which has a trigonometric function inside of it

Thanks for the reply. It helped more than 50% towards understanding all this (the crucial part being sin(x + 2*k*pi) = sin(x) because this explains why the right inverse has "+pi" in it).

It seems that I've left out a crucial detail in my question. The original problem was to determine the value for f^-1(x) when x = 20 and the answer given is 7 against which I've been checking my inverse function, although this wasn't the actual assignment (so what I'm trying to do here is being overly meticulous and actually derive an inverse function when in fact I could just refer to a graph of the original function): The way I did it in the beginning was This led me to believe that I can't just assume arcsin(sin(x)) = x unless sin(x) is an element of [-pi/2, pi/2], for reasons you've stated as well.

But this is as far as I got, actually, realizing that I don't have enough trigonometry in my head to figure out how this should be worked out. I still don't understand why it's -arcsin() instead of arcsin(), although I vaguely picture it having something to do with the fact that input value for example 20 would land outside [-pi/2, pi/2] and thus on "the wrong side" of the unit circle.

#### Search Tags

deriving, function, inside, inverse, inverse function, inversing, trigonometric 