Arccos(x+y)=? addition theorems for inverse trigonometric functions?

[raul_l]

Hi.
Are there any addition theorems for inverse trigonometric functions?
Like arccos(x+y)=? or something...

I was wondering about this when I tried to find the derivative of f(x)=arccos(x) by setting
f'(x)=\frac{\arccos(x+\Delta x)-\arccos(x)}{\Delta x}

 

[HallsofIvy]

No, there aren't. To find the derivative of arccos(x), use the fact that
if y= arccos(x) then x= cos(y) and the fact that
\frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}
(from the chain rule).

 

[Hurkyl]

Well, there are, but they're not always pretty.

E.G. there is a formula for:

cos u + cos v


now, if you let u = arccos x, v = arccos y, and take the arccos of both sides, you get something (ugly).

If I were to use trig identities for this, I'd be more likely to look at the cosine (or sine) of the entire numerator.


Of course, the "right" way to do it is as HoI suggested. But trig identities might be fun.

 

[raul_l]

cos u + cos v


now, if you let u = arccos x, v = arccos y, and take the arccos of both sides, you get something (ugly).

Not sure what you mean. Whatever I try, I get back to square one, basically proving that x+y=x+y.
Maybe you could explain it a little bit further? (or point me in the right direction)

 

[Werg22]

Well, there is certainly a way to express it as two different functions, one of x and one of y, using a Taylor's expansion... but that's quite useless.

 

[Matthew Rodman]

Ok, set

y = \arccos{(x)}

and

y^{\dagger} = \arccos{(x + \Delta x)}

now use the identity

\cos{u} - \cos{v} = -2 \sin{\frac{(u-v)}{2}} \sin{\frac{(u+v)}{2}

which gives us

\cos{y^{\dagger}} - \cos{y} = \Delta x = -2 \sin{\frac{(y^{\dagger}-y)}{2}} \sin{\frac{(y^{\dagger} + y)}{2}}

but for small delta x, we can approx for the sines

\Delta x \approx - ( y^{\dagger}-y) \sin{y}

or, in other words

y^{\dagger} - y \approx - \frac{\Delta x}{\sin{(\arccos{x})}} = -\frac{\Delta x}{\sqrt{1-x^2}}

hence

\frac{\arccos{(x+\Delta x)} - \arccos{(x)}}{\Delta x} \approx -\frac{\Delta x}{\Delta x \sqrt{1-x^2}}

which gives you your answer.

 

[Hurkyl]

but for small delta x, we can approx for the sines

It takes a bit of work to show that you really can do that, methinks.

 

[Matthew Rodman]

I'm not sure I understand - I'm only considering the case where delta x is small (as befits the question), so

y^{\dagger} - y

is small, so

2 \sin{(y^{\dagger} - y)/2} \approx y^{\dagger} - y

and also

\sin{(y^{\dagger} + y)/2} \approx \sin{(y)}

(or roughly sin (y^dagger)... it makes no odds)

could you elaborate?

edit: also, I've assumed that y is one of the many values produced by the arccos, and that y^dagger is the value that is delta-y away from it, and not a multiple of 2 pi.

 

[raul_l]

Makes sense to me.
Thanks.

edit:

It takes a bit of work to show that you really can do that, methinks.

Isn't it true that you can replace sine with its argument in the process, where the argument approaches zero?
According to
\lim_{x\rightarrow0} \frac{sinx}{x}=1
sine(x) and x are equivalent functions (I'm probably not using the right word here but hopefully you understand what I mean) and are interchangeable.

 

[Matthew Rodman]

Raul_l: I'm guessing that the issue concerns the multiple values that arccos returns. However, as long as you state the specifics of y and y^dagger as I've mentioned above (in the edit bit), it follows. I think if you are submitting this as a homework response, for example, one would need to be explicit about this.

If I'm wrong, then ignore this post.

 

[Hurkyl]

The main thing is that, while y'all are being fairly careful about everything else... you completely ignore the fact that you need to invoke the continuity of arccos to determine that the difference between the two y's goes to zero. The proof looks like:

{small step}
{small step}
{leap over an important issue}
{small step}
{small step}

and so it looks incomplete. I thought you might even need to invoke more, but I was wrong.


It also seemed you were glossing over some things involving the approximations, but after reviewing it, I see there aren't any technical faults there.

 

[Matthew Rodman]

Thanks, Hurkyl. You're point is a good one - brevity isn't always advisable.

 

[raul_l]

I'm guessing that the issue concerns the multiple values that arccos returns.

How does arccos return multiple values? Do you mean something like I have in the attachment?
The way I see it, if f is a function of arccos(x) then by definition for every element x there is no more than one f(x).
Keeping that in mind, I'm not sure where the continuity problem comes in..

 

[Matthew Rodman]

Because there are an infinitely many arguments to cos (separated by multiples of 2 pi) that produce the same cos value, for any given argument of arccos, there are an infinite number of return values (separated by multiples of 2 pi). Namely,

y_0 + 2 m \pi = \arccos{x} where m is any integer.

As long you're clear what you're doing, you should be ok. That's why I stipulated that the y^dagger value be the one which is Delta-y away from the y value - to avoid confusion.

 

[raul_l]

I understand why arccos returns an infinite number of values. I just thought that if we are talking about a function of arccos then the range of the function should be a fixed set of output values where the difference of two values is no more than \pi .
For example if the range is chosen to be [-\pi;0] then all values of a given function should fall into that range, including y and y^dagger in this case.
I'm probably talking nonsense here, but I find it hard to look past the fact that a function can at any point return only one value (by definition). Because if not then the graph of the function should be something like I have in the attachment of a previous post and that doesn't make sense at all (to me).

 

[Matthew Rodman]

I'm probably talking nonsense here, but I find it hard to look past the fact that a function can at any point return only one value (by definition).

Err... what about

f(x) = \sqrt{x} ?

I don't think I understand your problem, on the whole, but the assertion that you can treat any arbitrary function as returning only one value seems a bit suspect, if that's what it is you're trying to say.

So, any function of arccos must take into account that fact that it will, probably, have an infinite number of arguments, and therefore will likely have a multi-valued output.

If you start imposing a single-valued nature on a function, you'll be creating problems for yourself. (Of course, if I've misread your comment, forget this).

 

[raul_l]

What I meant was that every argument produces only one value that has to fall into the range.
By "at any point" I meant that if we arbitrarily take a point on the x-axis then there is only one corresponding point on the y-axis.

 

[Matthew Rodman]

Ok, yes, which is no less than what I did earlier with arccos.