Trig identity with natural logs and absolute value?

[jhahler]

Trig identity with natural logs and absolute value??

Homework Statement

-ln|csc(x) + cot(x)|= ln|cscx(x)-cot(x)|

Homework Equations

The Attempt at a Solution

I got that csc(x)=1/sin(x) and cot(x)=cos(x)/sin(x), giving me a common denominator, added together I have 1+cos(x)/sin(x). So now the Absolute value and ln are throwing me a curve, and I'm stuck at -ln|1+cos(x)/sin(x)| = ln|1-cos(x)/sin(x). Even if you can just point me in the right direction it would be a big help, thanks in advance!

 

[Bread18]

Try to get rid of the ln.

 

[conquest]

You should be able to get rid of the absolute value signs partly by just looking at intervals 0<=x<=1/2 pi etc. But it seems like there will be trouble since the argument of the logarithm vanishes at some points probably worth a note. you have -ln at one side maybe you should taking the inverse of the argument there.

 

[conquest]

Try to get rid of the ln.

Haha right or just do that

 

[jhahler]

how do you get rid of the ln? does it have to do with multiplying by the inverse?

 

[Bread18]

Also remember the rules for logarithms.

$lnx^{k}=klnx$

 

[jhahler]

not sure how to relate lnx^k to this problem.. is x the function? what is k?

 

[Bread18]

not sure how to relate lnx^k to this problem.. is x the function? what is k?

I meant it as a general rule for logarithms. In this problem I mean.

$-ln|cosecx + cotx| = ln(cosecx + cotx)^{-1}$

 

[Curious3141]

Homework Statement

-ln|csc(x) + cot(x)|= ln|cscx(x)-cot(x)|

Homework Equations

The Attempt at a Solution

I got that csc(x)=1/sin(x) and cot(x)=cos(x)/sin(x), giving me a common denominator, added together I have 1+cos(x)/sin(x). So now the Absolute value and ln are throwing me a curve, and I'm stuck at -ln|1+cos(x)/sin(x)| = ln|1-cos(x)/sin(x). Even if you can just point me in the right direction it would be a big help, thanks in advance!

Hints:

$\ln x^k = k\ln x$. What happens when $k = -1$?

$\frac{a}{b+c} = \frac{(a)(b-c)}{(b+c)(b-c)}$. This form will come in useful. Simplify the denominator and see how you can apply a trig identity to make it really simple.

 

[jhahler]

when k = -1 does that mean take the inverse, like ln|1+cos(x)/sin(x)|^-1 ?

 

[Curious3141]

when k = -1 does that mean take the inverse, like ln|1+cos(x)/sin(x)|^-1 ?

Yes. Remember it's the logarithm of the inverse, NOT the inverse of the whole logarithm.

It's clearer if you express the inverse as 1/(something). Leave everything in terms of cot and csc, and apply the second hint I gave you.

 

[Bread18]

$ln(cosecx + cotx)^{-1} = ln|\frac{1}{cosecx + cotx}|$

So now you know that

$ln|\frac{1}{cosecx + cotx}| = ln|cosecx - cotx|$

 

[Curious3141]

$ln(cosecx + cotx)^{-1} = ln|\frac{1}{cosecx + cotx}|$

So now you know that

$ln|\frac{1}{cosecx + cotx}| = ln|cosecx - cotx|$

Bread, it's not an equation he's supposed be solving. He's supposed to prove a trig identity. So he's supposed to manipulate one side (in this case, starting from the LHS is fine) till it equals the other. He's not supposed to start by assuming equality.

 

[Bread18]

Bread, it's not an equation he's supposed be solving. He's supposed to prove a trig identity. So he's supposed to manipulate one side (in this case, starting from the LHS is fine) till it equals the other. He's not supposed to start by assuming equality.

Ah k, I wasn't too sure what he was trying to do, he didn't really state what the question said.

Though I do now have a question about that, if I have a question that says show a=b, can't i do so by showing that they both = c?

 

[jhahler]

OK! got it now, the -ln makes me take the inverse by raising the power to negative 1, then I put 1/csc(x)+cot(x) then the 1 on top goes away when I switch the sign on the bottom to negative. Awesome! So i didn't need to worry about the absolute value, or figuring out that csc(x)+cot(x) equaled 1 + cos(x)/sin(x). Thanks again!

 

[Curious3141]

Ah k, I wasn't too sure what he was trying to do, he didn't really state what the question said.

Though I do now have a question about that, if I have a question that says show a=b, can't i do so by showing that they both = c?

It's in the thread title.

You can do that, but it's usually preferable to try to manipulate one side until it becomes the other.

Sometimes, to "see" the solution, you use the method you alluded to (to simplify both to the same form), but once you do this, you can reverse engineer one side of it so that you can work from LHS to RHS or vice versa when you present the final proof.

 

[Curious3141]

OK! got it now, the -ln makes me take the inverse by raising the power to negative 1, then I put 1/csc(x)+cot(x) then the 1 on top goes away when I switch the sign on the bottom to negative. Awesome! So i didn't need to worry about the absolute value, or figuring out that csc(x)+cot(x) equaled 1 + cos(x)/sin(x). Thanks again!

Not so fast. You'd better show exactly what you did, because something about what you wrote doesn't seem right.

 

[Bread18]

OK! got it now, the -ln makes me take the inverse by raising the power to negative 1, then I put 1/csc(x)+cot(x) then the 1 on top goes away when I switch the sign on the bottom to negative. Awesome! So i didn't need to worry about the absolute value, or figuring out that csc(x)+cot(x) equaled 1 + cos(x)/sin(x). Thanks again!

What exactly did you do?

 

[Bread18]

It's in the thread title.

You can do that, but it's usually preferable to try to manipulate one side until it becomes the other.

Sometimes, to "see" the solution, you use the method you alluded to (to simplify both to the same form), but once you do this, you can reverse engineer one side of it so that you can work from LHS to RHS or vice versa when you present the final proof.

I generally work until they reach a common solution, like this I showed they were both equal to $sin^{2}x$

 

[Curious3141]

I generally work until they reach a common solution, like this I showed they were both equal to $sin^{2}x$

Actually, there's a much more direct way here. Can you work it out with the hints I provided?

BTW, I don't think either side is equal to $\sin^2{x}$.

 

[Bread18]

Actually, there's a much more direct way here. Can you work it out with the hints I provided?

I have done it another way, I think it's what he is saying.

$\frac{1}{cosecx + cotx}\times\frac{cosecx - cotx}{cosecx - cotx}$

$= \frac{cosecx - cotx}{\frac{sin^{2}x}{sin^{2}x}}$

$=cosecx - cotx$

 

[Curious3141]

I have done it another way, I think it's what he is saying.

$\frac{1}{cosecx + cotx}\times\frac{cosecx - cotx}{cosecx - cotx}$

$= \frac{cosecx - cotx}{\frac{sin^{2}x}{sin^{2}x}}$

$=cosecx - cotx$

Yup, that's right. For the denominator, it's acceptable to recognise the trig identity $\csc^2{x} - \cot^2{x} = 1$ directly. But it would've been good if the OP had worked this out for himself. Right now, neither of us is sure what he did.

 

[jhahler]

ok guys, sorry, the inverse of csc(x)+cot(x) is 1/csc(x)+cot(x), and I saw the 1 on top and just thought it meant switch csc(x)+cot(x) to csc(x)-cot(x), making it equal to the other side, proving the identity, isn't that right?

 

[Curious3141]

ok guys, sorry, the inverse of csc(x)+cot(x) is 1/csc(x)+cot(x), and I saw the 1 on top and just thought it meant switch csc(x)+cot(x) to csc(x)-cot(x), making it equal to the other side, proving the identity, isn't that right?

Good thing you posted.

A couple of things:

1) Brackets are important. $\frac{1}{a} + b \neq \frac{1}{a+b}$. When you write "1/csc(x)+cot(x)", it's generally read as the former, while you meant the latter. So please be careful about placing your brackets to render the right expression.

2) In general, $\frac{1}{a+b} \neq a - b$! It's true in this case because of the particular nature of those trig ratios, but you have to *show* that it's true in this particular case. But to see why it isn't true in general, try a = 3, b = 2.

 

[jhahler]

ok, yeah I need to learn how you guys post on here in math terms instead of the keyboard, sorry for the confusion of 1/csc(x)+cot(x) being read as (1/csc(x))(cot(x)).. and yeah I just got lucky, because I didn't even think to multiply the top and bottom by the denominator. Thanks again! P.S. is it pushing it to post twice in 1 night? I got 1 more problem I need answered but I feel like that's frowned on around here, thanks again!

 

[Bread18]

ok, yeah I need to learn how you guys post on here in math terms instead of the keyboard, sorry for the confusion of 1/csc(x)+cot(x) being read as (1/csc(x))(cot(x)).. and yeah I just got lucky, because I didn't even think to multiply the top and bottom by the denominator. Thanks again! P.S. is it pushing it to post twice in 1 night? I got 1 more problem I need answered but I feel like that's frowned on around here, thanks again!

Ask anything you need help with. This is a forum to help people.

 

[Mentallic]

ok, yeah I need to learn how you guys post on here in math terms instead of the keyboard, sorry for the confusion of 1/csc(x)+cot(x) being read as (1/csc(x))(cot(x))..

Actually it would rather be 1/(csc(x)+cot(x)) and the way you right it with Latex is by typing

[ tex]\frac{1}{\csc(x)+\cot(x)}[ /tex] (without the spaces in the tex tags)

is it pushing it to post twice in 1 night? I got 1 more problem I need answered but I feel like that's frowned on around here, thanks again!

Haha where did you get that idea? Post away!

 

[jhahler]

Ok! thanks Mentallic, Already posted second question, now I'm going to have to start using Latex.