cos(infinity) = sin(infinity)?

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.

Re: cos(infinity) = sin(infinity)?

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

$\displaystyle f(\infty) = \lim_{x\to\infty} f(x)$, 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.

Re: cos(infinity) = sin(infinity)?

Quote:

Originally Posted by

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

$\displaystyle f(\infty) = \lim_{x\to\infty} f(x)$, 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

Re: cos(infinity) = sin(infinity)?

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.

Re: cos(infinity) = sin(infinity)?

Quote:

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 $\displaystyle 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 $\displaystyle \boxed{\Sigma}$ clicking on that give the LaTeX wraps, [tex] [/tex]. The code goes between them.

Quote:

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 $\displaystyle a_n=2n\pi$ and $\displaystyle b_n=(2n-1)\pi$ then $\displaystyle \cos(a_n)\to 1$ and$\displaystyle \cos (b_n)\to -1$.

That makes your idea impossible.

Re: cos(infinity) = sin(infinity)?

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).