cos(infinity) = sin(infinity)?

**AlderDragon** · March 28th 2011, 05:01 PM

I am using a TI-89 calculator. The highest level of math I've taken is high school calculus. I was playing around with infinity, and entered cos(infinity), which returned sin(infinity). Why is this? I don't understand limits incredibly well, but I know that the range of the sine and cosine functions fluctuates between 1 and -1. I guess you could also say that their domain is the same, as -infinity to infinity? It is approaching infinity along the x axis, but does not approach anything along the y-axis?

Can anyone shed more light on this? Thanks.

**Plato** · March 28th 2011, 05:11 PM

May I tell you that $\infty$ is not a real number.
That is a shock to many. I hope you are not among them.
Therefore neither $\sin(\infty)$ nor $\cos(\infty)$ has any meaning.
If they are meaningless, then there is no comparison. Is there?

**bkarpuz** · March 28th 2011, 05:11 PM

You should not trust your calculators unless you do 4 basic operations.

**Drexel28** · March 28th 2011, 05:27 PM

Originally Posted by AlderDragon

I am using a TI-89 calculator. The highest level of math I've taken is high school calculus. I was playing around with infinity, and entered cos(infinity), which returned sin(infinity). Why is this? I don't understand limits incredibly well, but I know that the range of the sine and cosine functions fluctuates between 1 and -1. I guess you could also say that their domain is the same, as -infinity to infinity? It is approaching infinity along the x axis, but does not approach anything along the y-axis?

Can anyone shed more light on this? Thanks.

If I had to guess at the meaning that the calculator attributed to this I would guess that that it made some calculation like $\cos(\infty)=\cos(-\infty)=\cos\left(\frac{\pi}{2}-\infty\right)=\sin(\infty)$

**AlderDragon** · March 28th 2011, 05:27 PM

I realize that infinity isn't a real number, but I'm not convinced that they have no meaning. If there is no comparison then why does my calculator claim they are equivalent?

**topsquark** · March 28th 2011, 05:32 PM

Originally Posted by AlderDragon

I realize that infinity isn't a real number, but I'm not convinced that they have no meaning. If there is no comparison then why does my calculator claim they are equivalent?

(shrugs) My TI-92 occasionally tells me that $10^{-14} = 0$ . Your's is a more specific glitch but please note that my calculator (which is essentially the same as yours in most details) reports that $cos( \infty ) = cos( \infty )$ . It's just a programming glitch of some kind.

-Dan

**Plato** · March 28th 2011, 05:36 PM

Originally Posted by AlderDragon

I realize that infinity isn't a real number, but I'm not convinced that they have no meaning. If there is no comparison then why does my calculator claim they are equivalent?

Because your calculator is a dumb machine.
It is only as good as the person who programed it.
The $cos(x)$ can be any number if $x$ is very large.

**AlderDragon** · March 28th 2011, 05:46 PM

So ultimately it is explained by the fallibility of the calculator. That's fine - I just thought it might have something to do with the properties of sine and cosine.

**Corpsecreate** · June 4th 2011, 09:35 AM

Originally Posted by Plato

Because your calculator is a dumb machine.
It is only as good as the person who programed it.
The $cos(x)$ can be any number if $x$ is very large.

I think you mean it can be any number between and including -1 to 1.

As previously said, infinity is not a number so it cant be treated as one. Its like saying, what value on the real number line corresponds to 5*bedroom? The question doesnt make sense, so anything the calculator may spit out is worthless information.

**Deveno** · June 14th 2011, 09:12 PM

well, in general, f(∞) may not be meaningless. in the case of real (or even rational) numbers, we can set:

$f(\infty) = \lim_{x\to\infty} f(x)$ , IF this limit exists (well, technically, we should also have some other restrictions on f, but

normally continuity will suffice).

both the sine and cosine function oscillate. however, the "phase shift" between sine and cosine remains constant, no matter how large x becomes:

as functions they do not "approach each other". since calculators are necessarily finite and have a limit to how large a number they can store,

(which typically is actually an integer (my guess is 1 less than some power of 2), converted to scientific notation when too large to display),

my guess is that "default values" for f(x), when x exceeds the maximum value storeable, are pre-programmed in. most of these are probably not to be trusted.

**CaptainBlack** · June 15th 2011, 02:26 AM

Originally Posted by Deveno

well, in general, f(∞) may not be meaningless. in the case of real (or even rational) numbers, we can set:

$f(\infty) = \lim_{x\to\infty} f(x)$ , IF this limit exists (well, technically, we should also have some other restrictions on f, but

normally continuity will suffice).

both the sine and cosine function oscillate. however, the "phase shift" between sine and cosine remains constant, no matter how large x becomes:

as functions they do not "approach each other". since calculators are necessarily finite and have a limit to how large a number they can store,

(which typically is actually an integer (my guess is 1 less than some power of 2), converted to scientific notation when too large to display),

my guess is that "default values" for f(x), when x exceeds the maximum value storeable, are pre-programmed in. most of these are probably not to be trusted.

Calculators operate with a peculiar system of numbers, most of the peculiarities are not relevant here but what is (in modern systems anyway) are the ideal elements +inf, -inf, nan. The first denotes anything greater that the greatest representable number, the second anything less than the smallest representable number, and the third anything which is undefined (it stands for "not a number").

(For consistency 1/0 might return +inf rather than nan etc.)

Now if your calculator is programmed correctly and it allows you to enter this nonsense, you should get

cos(inf)=nan.

But note the operative word here is nonsense.

CB

**cheesehead** · December 4th 2012, 11:47 AM

Can one just define cos(\infinity) to be 0 via the following limit process? \int_0^\infty sin(x) dx=1-cos(\infty)=1, where cos(\infty)=0. The integral was done by adding a factor e^(-\alpha x) to the integrand and letting \alpha-> 0. Not sure how to type equations, so I wrote it in Tex format.

**Plato** · December 4th 2012, 12:17 PM

Originally Posted by cheesehead

Not sure how to type equations, so I wrote it in Tex format.

Here is how one uses LaTeX.
[tex] x = \neg A\cdot \neg B\cdot C + \neg A\cdot B\cdot \neg C + \neg A\cdot B\cdot C + A\cdot \neg B\cdot \neg C [/tex] gives $x = \neg A\cdot \neg B\cdot C + \neg A\cdot B\cdot \neg C + \neg A\cdot B\cdot C + A\cdot \neg B\cdot \neg C$
Click on the “go advanced” tab. On the toolbar you will see $\boxed{\Sigma}$ clicking on that give the LaTeX wraps, [tex] [/tex]. The code goes between them.

Originally Posted by cheesehead

Can one just define cos(\infinity) to be 0 via the following limit process?

No! the cosine function is continuous.

Define $a_n=2n\pi$ and $b_n=(2n-1)\pi$ then $\cos(a_n)\to 1$ and $\cos (b_n)\to -1$ .
That makes your idea impossible.

**Deveno** · December 4th 2012, 02:18 PM

as x goes to infinity, cos(x) takes on EVERY VALUE between -1 and 1.

this is a bit easier to understand if you look at the behavior of cos(1/x) near 0 (this switches the "far ends" of the real line to a bounded area we can look at in detail). there's no number we can pick in the interval [-1,1] that will make cos(1/x) continuous at 0 (it's oscillating too quickly).

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