# Canceling Sin's

Dumb moment here...I have the following equation:

Sin (4/(3x+3)) / Sin (4/3x)= 1

can i cancel out the sin's?

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rock.freak667
Homework Helper
Dumb moment here...I have the following equation:

Sin (4/(3x+3)) / Sin (4/3x)= 1

can i cancel out the sin's?
No you can't just randomly cancel it out.

But you can rearrange and say that if sinX=sinY then it implies that X=Y.

what if i was taking the limit of that equation:

Lim Sin (4/(3x+3)) / Sin (4/3x)
x->inf

rock.freak667
Homework Helper
I think the lim sinf(x) = sin lim f(x) is valid if I remember correctly.

Mark44
Mentor
Dumb moment here...I have the following equation:

Sin (4/(3x+3)) / Sin (4/3x)= 1

can i cancel out the sin's?
As long as a isn't 0 you can do this:
$$\frac{a\cdot b}{a \cdot c} = \frac{b}{c}$$

Do you notice a significant difference between what I did compared to what you're trying to do?

Dumb moment here...I have the following equation:

Sin (4/(3x+3)) / Sin (4/3x)= 1

can i cancel out the sin's?
Isn't this equivalent to

working on it-

Last edited:
No, you can't cancel the sins!!!! Imagine this equation:

$${\sin(x)\over\sin(y)}=2$$

If you cancel the sin's you get $$x/y=2$$, which is wrong. But what about taking the denominator to the other side? Then you can do something clever: if $$sin(x)=sin(y)$$, then there is a known relation between x and y. Either they're equal or...

Dumb moment here...I have the following equation:

Sin (4/(3x+3)) / Sin (4/3x)= 1

can i cancel out the sin's?
The letters s-i-n in $\sin(x)$ are not variables. Those three letters together stand for an operation -- namely the operation of computing the sine of x. Similarly, we use the "+" symbol to refer to the operation of adding two values.

"Canceling" is the notion of dividing out by a common nonzero number, or by a common variable that stands for a nonzero number. "sin" is not a variable; it is an operation. Canceling is not a matter of "deleting letters and symbols" that appear in both the numerator and denominator.

Mark44
Mentor
The letters s-i-n in $\sin(x)$ are not variables. Those three letters together stand for an operation -- namely the operation of computing the sine of x. Similarly, we use the "+" symbol to refer to the operation of adding two values.

"Canceling" is the notion of dividing out by a common nonzero number, or by a common variable that stands for a nonzero number. "sin" is not a variable; it is an operation. Canceling is not a matter of "deleting letters and symbols" that appear in both the numerator and denominator.
Does this mean I can't do this...
$$\frac{sin x}{n} = 6$$
? ?

I thought that $$i=\sqrt{-1}$$ I'm confused....:)

Yeah, so funny... ;)

But if you want the limit, still can't cancel the sin's (cancel the sins... what would a priest think?) :P But you can do the obvious thing, try to substitute. You get sin(0)/sin(0) so, 0/0... why don't you try now L'Hôpital?

Char. Limit
Gold Member
I'd really like someone to cancel my sins...

Yeah, so funny... ;)

But if you want the limit, still can't cancel the sin's (cancel the sins... what would a priest think?) :P But you can do the obvious thing, try to substitute. You get sin(0)/sin(0) so, 0/0... why don't you try now L'Hôpital?

http://en.wikipedia.org/wiki/L'Hôpital's_rule

which gives some good examples on howto use L'Hospitals rule. Something which is a bit strange is that why is it to become a famous mathematician you have to so strange names? ;)

Only mathematician with a easy name to remember is Niels Henrik Abel and Cauchy.

:D Susanne

Jrlaguna's conjecture: the higher you climb into advanced mathematics, the more strange the names of the theorems will get. A proof is required, but we have good evidence of that... what do you think of "Gupta-Bleuler quantization"?

Char. Limit
Gold Member

http://en.wikipedia.org/wiki/L'Hôpital's_rule

which gives some good examples on howto use L'Hospitals rule. Something which is a bit strange is that why is it to become a famous mathematician you have to so strange names? ;)

Only mathematician with a easy name to remember is Niels Henrik Abel and Cauchy.

:D Susanne
I'd also include Euclid, Newton, Leibniz, and Neumann on that list, but in general I think you're right. I still can't remember how to spell Ramananan, and he seems to have done everything he can with numbers.

I'd also include Euclid, Newton, Leibniz, and Neumann on that list, but in general I think you're right. I still can't remember how to spell Ramananan, and he seems to have done everything he can with numbers.

Euclid, Newton off cause... :D If I invent a some theory which actually works one day, then I will change my name to L'Susanniwitz then I will surely be remembered :)

You know the sense of humour of physicists... so when you come up with an important equation, this thread will come up and they will call it the Susanniwitz equation, no matter what your wishes are by then! Sorry, darling, you're dooooomed! :) :) :)

You know the sense of humour of physicists... so when you come up with an important equation, this thread will come up and they will call it the Susanniwitz equation, no matter what your wishes are by then! Sorry, darling, you're dooooomed! :) :) :)
Yeah you are right.

By the way back to the original problem and by using L'Hospital one runs into one of funny requirements that in order for that old french guys theory to work then the limit as show in the example has to be there. And I surgest to the original poster to test if the limit

$$\lim_{x \to \infty} \frac{f(x)}{g(x)}$$ exists Yeah, funny as it is the other discussion, let us focus, boys and girls, ok? :)

In this case, the limit does exist, the conditions on the theorem are met. Promised. The question poser has to work it out, though. (This is homework help.)

Yeah, funny as it is the other discussion, let us focus, boys and girls, ok? :)

In this case, the limit does exist, the conditions on the theorem are met. Promised. The question poser has to work it out, though. (This is homework help.)
The funny thing is I tested this problem on the equation-solver on old TI-92 and it claims that there are several solutions to the OP problem like the solutions are polynomial...

You mean the eq, sin(...)/sin(...)=1, right? The limit has a unique solution, and he can find it with... L'H. :P

Cyosis
Homework Helper
The funny thing is I tested this problem on the equation-solver on old TI-92 and it claims that there are several solutions to the OP problem like the solutions are polynomial...
Your calculator is correct. Don't forget that if you have an equation of the type sin(x)=sin(y) then x=y+2pi*k is a solution for k an integer.

Your calculator is correct. Don't forget that if you have an equation of the type sin(x)=sin(y) then x=y+2pi*k is a solution for k an integer.
Okay,

I guess even though I have dropped my TI Calculator in the floor, spilled Coca Cola on it, hell it still works :D