# Absolute value confusion

1. Mar 30, 2009

### deancodemo

1. The problem statement, all variables and given/known data
Find $$\frac{d}{dx} \sin^{-1}(\sin x)$$

2. Relevant equations

3. The attempt at a solution
Now, I know that the above expression simplifies directly to $$\frac{d}{dx} x = 1$$, but I attempted it the long way. Here it is:
$$\frac{d}{dx} \sin^{-1}(\sin x)$$

$$= \frac{\cos x}{\sqrt{1 - \sin^2 x}}$$

$$= \frac{\cos x}{\sqrt{\cos^2 x}}$$

$$= \frac{\cos x}{|\cos x|}$$

$$= 1 \mbox{ if } \cos x > 0$$

$$= -1 \mbox{ if } \cos x < 0$$

I know that this is incorrect! The answer should be 1. Did I incorrectly use the definition of $$\sqrt{x^2} = |x|$$?

2. Mar 30, 2009

### sutupidmath

do you have any restrictions on the domain of your function?

3. Mar 30, 2009

### deancodemo

No. Well, what I mean is that you can put in any value of x. Then -1 <= sinx <= 1 which is the desired domain for arcsin x, then arcsin x produces values such that -pi/2 <= arcsin x <= pi/2.

4. Mar 30, 2009

### sutupidmath

Ok. We know that the inverse of sinx with which we usually work is defined on the interval [-pi/2,pi/2]. So, in this case, we are basically working with values of x on this interval, where cosx is always positive.So, basically we can put in only values of x from -pi/2 to pi/2, and cos x for all these values is always positive, so you never get cosx<0 for these values of x.

Last edited: Mar 30, 2009
5. Mar 31, 2009

### meiso

Your derivation the "long way" was actually completely correct, as was your usage of absolute value in taking the square root of a squared real variable.

It is your assumption that sin-1(sin(x)) = x that is incorrect. (please excuse my lack of formatting; sin-1(x) is sine inverse)
f-1(f(x)) = x assuming that f-1(x) is the true inverse of f.

sin-1(x) is given a domain restriction ( -pi/2 <= x <= pi/2 ) because the actual inverse of sin(x) is not a function, stemming from the fact that sin(x) is not a one-to-one function. So, sin-1(x) is the inverse of just a part of the sine function.

Before we even involve any calculus, you can see that sin-1(sin(x)) is not always true with this counterexample:

sin-1(sin(3pi/2)) = -pi/2

Now, while 3pi/2 and -pi/2 are of course two ways to state the measure of the same angle, they are not the same value (obviously, 4.712... is not equal to -1.571...)

So, the expression sin-1(sin(x)), using sin-1(x) as it is normally defined, cannot be simplified to x.

Going back to the result of your derivation, an expression like cos(x)/abs(cos(x)) can be thought of as the function sign(cos(x)), except when cos(x)=0. So you are in fact correct in your inequality statements at the end of your posting.

Now, note that,

cos(x) > 0 when -pi/2 + 2pi(n) < x < pi/2 + 2pi(n) where n is an integer

cos(x) < 0 when pi/2 + 2pi(n) < x < -pi/2 + 2pi(n) where n is an integer

If you graph the sin-1(sin(x)) function, you will see that it is a jagged-looking periodic function with an amplitude of pi/2 (expected from the range of sin-1(x)) that is basically a piecewise function constructed of lines with the slope of 1 and -1 in alternating intervals, corresponding to the sign of the cosine function at each interval. So, looking at the definition of the derivative as the slope of a function at a point, it fits with your conclusion perfectly.

Notice that if cos(x) = 0 (which happens when x is equal to odd multiples of pi/2), the expression you found seems that it would be undefined. That is also a correct result, and you would also notice on the graph of sin-1(sin(x)) that the function is not differentiable at the sharp turns that occur when x is equal to an odd multiple of pi/2.

As sutupidmath said, if you restrict the domain of the function to -pi/2 < x < pi/2, cos(x) will always be positive, then sin-1(sin(x)) will be equal to x, and the derivative will be 1. But if you want to take the derivate for the function sin-1(sin(x)) with a domain of all real numbers (the same domain as the inner function sin(x)), then your way, once again, is correct, with the added restriction at the end that the derivative is undefined at odd multiples of pi/2.

Last edited: Mar 31, 2009
6. Apr 1, 2009

### deancodemo

meiso, what you said was completely correct. I now understand where I went wrong.

You cannot simply deduce sin-1(sinx) = x for any value of x that you put into sin-1(sinx).
The domain of sinx was restriced to -pi/2 <= x <= pi/2 so that the function defined on this interval is monotonic increasing and hence we can find the inverse function in this domain. This means that sin-1 x has a range equal to the restricted domain of sinx (-pi/2 <= x <= pi/2).
I was getting confused was because I thought that sin-1(sinx) = x and hence the gradient of sin-1(sinx) is always 1. But it isn't.

Here is the graph of sin-1 x with alternating gradients at the intervals you specified: