# Stupid trig simplifying

1. Oct 13, 2008

### lax1113

1. The problem statement, all variables and given/known data
(secx-1)/x^2 limit x--->0

2. Relevant equations
secx=1/cosx
(1-cosx)/x=0

3. The attempt at a solution
I have done obvious thing, solved to make secent cosine and got common denominators in the top portion of the fraction, then subtracted. Ended up with {(1-cosx)/cosx}/x^2.
So when x-->0 its still 0/0 which isn't good. I'm pretty sure that i have to somehow get it to be (1-cosx)/x because we learned that identity in calc a while back, and I am not sure if i just dont remember something from that to apply to this physics, or if I am approaching it all wrong.

2. Oct 13, 2008

### CompuChip

What did you learn about (1 - cos(x))/x? Maybe you can apply the same procedure here?

3. Oct 13, 2008

### krausr79

L'hopital?

If this is calculus then you can use l'hopitial's rule which says if lim of something is in the form 0/0 then take the derivative of the top and bottom and do the limit of that.

derivative of 1/cos-1 is sin/cos2
derivative of x2 is 2x

still 0/0 so do it again

derivative of sin/cos2 is 1/cos + 2sin2/cos3
derivative of 2x is 2

now it's 1/2 flat
also, my TI-92 says the limit is 1/2

http://en.wikipedia.org/wiki/LHopitals_rule

4. Oct 13, 2008

### lax1113

Krausr,
thank you, it is in fact 1/2, says the back of the book. However, I am not sure if i should even right what you just put because thers no way in hell i could have fallen asleep for that long during class . That was quite an answer, thank you very much krausr, i'm going to reread it a coupel times and see if it rings a bell but it seems like something we have yet to do.... Maybe thats the point?

I hate when teachers do that, throw a question at you that is unsolvable just to get you to think!

Thanks again

5. Oct 13, 2008

### lax1113

Ok so i was actually trying something and I would love if someone could just pop in and say if this is right/at least makes sense, doens't violate any rules of algebra/trig.

(secx-1)/x^2

simplified using just common denom/sec identity to...

{(1-cosx)/(cosx)}/x^2 then...

{(1-cosx^2)/(cosx+cosx^2)}/x^2

using identities...

{(sinx^2)/(cosx(1+cosx))}/x^2

finally....

(sinx^2)/(cosx(1+cosx))x^2

so the sin^2 and the x^2 in the denominator can simplify to 1, because of the squeeze therom i believe....

so then it would be 1/(cosx(1+cosx)) cosx---x-->0

1/2

6. Oct 13, 2008

### CompuChip

How did you go from
{(1-cosx)/(cosx)}
to
{(1-cosx^2)/(cosx+cosx^2)}
?

I don't think this is a valid step. However, it brings me to an idea: maybe multiplying the first expression by (1 + cos x) / (1 + cos x) would work?

7. Oct 13, 2008

### lax1113

Did you by any chance attempt the step you told me to do? If you do follow through with that step, you might find the answer to be kinda familiar....

That is how i got to the 1/cosx^2.

8. Oct 14, 2008

### CompuChip

Hmm, what was I thinking that I got something else? Maybe it was just too late in the evening.

Never mind my last post then, it looks like you did it right. In an official proof, you might also want to show that the conditions of the theorem you are using are fulfilled, but otherwise it looks fine.

9. Oct 14, 2008

### lax1113

I did not mean to seem ungreatful or snobbish by the way compuchip, I just kinda thought it was funny. I do really appreciate the help, just happened to be a second too late. And yes, I know the feeling all too well of doing math too late, or on not enough sleep. I answered about half my physics questions with 32 feet/second^2 for freefall, while the entire test was clearly in meters :yuck:

10. Oct 15, 2008

### CompuChip

Don't apologise, it wasn't taken as ungreatful, snobbish or anything like that. I just didn't get (1 - cos x)(1 + cos x) to be equal to 1 - cos^2(x), but the next morning it suddenly turned out to be so

Ah well, even the best (cough!) make a little error sometimes :tongue: