# Simplifying a log expression with identities

Gold Member
I was supposed to simplify the expression ##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|## and apparently it’s wrong. Where’s the mistake? Is it not simplified enough or . . . ?

##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|##

##=\ln |\frac {\cos {x}}{\sin {x}}|+\ln |\frac {\sin {x}}{\cos {x}}\cdot \cos {x}|##

##=\ln |\frac {\cos {x}}{\sin {x}}|+\ln |\sin {x}|##

##=\ln |\frac {\cos {x}\sin {x}}{\sin {x}}|##

##=\ln |\cos {x}|##

If it’s really obvious, I’ll be happy with a hint. I thought that this was easy, but apparently I was mistaken?

member 587159
I was supposed to simplify the expression ##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|## and apparently it’s wrong. Where’s the mistake? Is it not simplified enough or . . . ?

##\ln |\cot {x}|+\ln |\tan {x}\cos {x}|##

##=\ln |\frac {\cos {x}}{\sin {x}}|+\ln |\frac {\sin {x}}{\cos {x}}\cdot \cos {x}|##

##=\ln |\frac {\cos {x}}{\sin {x}}|+\ln |\sin {x}|##

##=\ln |\frac {\cos {x}\sin {x}}{\sin {x}}|##

##=\ln |\cos {x}|##

If it’s really obvious, I’ll be happy with a hint. I thought that this was easy, but apparently I was mistaken?

Looks correct though. Note that Ln0 is undefined, so you might want to specify when the equality is correct.

Gold Member
Looks correct though. Note that Ln0 is undefined, so you might want to specify when the equality is correct.
Well, I was given the problem with no specifications and would it really affect the answer? I mean, this expression doesn’t deal with ##\ln {0}##.

Gold Member
So what is the error?

SteamKing
Staff Emeritus
Homework Helper
Well, I was given the problem with no specifications and would it really affect the answer? I mean, this expression doesn’t deal with ##\ln {0}##.
You sure about that? You mean cos x is never equal to zero?

• ProfuselyQuarky
Gold Member
You sure about that? You mean cos x is never equal to zero?
Oh, I never said cos x never is equal to zero. ##\cos {\frac {\pi}{2}}=0## and ##\cos {\frac {3\pi}{2}}=0## and so on.

Gold Member
But would that affect the answer? How did I simplify wrong?

SteamKing
Staff Emeritus
Homework Helper
Oh, I never said cos x never is equal to zero. ##\cos {\frac {\pi}{2}}=0## and ##\cos {\frac {3\pi}{2}}=0## and so on.
So what happens to ln |cos x| when x = (2k+1)π/2 ?

Gold Member
So what happens to ln |cos x| when x = (2k+1)π/2 ?
We get ln 0 which is undefined

Gold Member
Oh, wait, so I'd have to specify that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.

That's great but there wasn't a single example of any of that in my book which is why I was thinking that my error had something to do with using the identities wrong.

member 587159
Oh, wait, so I'd have to specify that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.

That's great but there wasn't a single example of any of that in my book which is why I was thinking that my error had something to do with using the identities wrong.

The answer is correct, as long as you specify for which values the expression is undefined.

• ProfuselyQuarky
SammyS
Staff Emeritus
Homework Helper
Gold Member
Oh, I never said cos x never is equal to zero. ##\cos {\frac {\pi}{2}}=0## and ##\cos {\frac {3\pi}{2}}=0## and so on.
Frankly. I don't think this is the problem. After all, the original expression isn't defined for values of x which make cos(x)=0, either.

However, there are values of x for which the original expression is undefined, but are defined for ln(cos(x)) . Throw those out.

• ProfuselyQuarky
member 587159
Why do you think your expression is wrong in the first place?

Gold Member
Why do you think your expression is wrong in the first place?
It was marked wrong.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Oh, wait, so I'd have to specify that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.

That's great but there wasn't a single example of any of that in my book which is why I was thinking that my error had something to do with using the identities wrong.
I'm not sure I was very clear in my previous post.

I don't think the problem is:
that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.​
Those values of x at which ##\cos {x}=0## are not in the domain of ## \ \ln |\cot {x}|+\ln |\tan {x}\cos {x}|\ ## in the first place.

I think that the problem is that ##\ln |\cos {x}|## is defined for some values of x for which ## \ \ln |\cot {x}|+\ln |\tan {x}\cos {x}|\ ## is not defined. I think that you must restrict the domain of the answer to eliminate those.

member 587159
I'm not sure I was very clear in my previous post.

I don't think the problem is:
that ##\ln |\cos {x}|## is only the answer when ##\cos {x}\neq0##.​
Those values of x at which ##\cos {x}=0## are not in the domain of ## \ \ln |\cot {x}|+\ln |\tan {x}\cos {x}|\ ## in the first place.

I think that the problem is that ##\ln |\cos {x}|## is defined for some values of x for which ## \ \ln |\cot {x}|+\ln |\tan {x}\cos {x}|\ ## is not defined. I think that you must restrict the domain of the answer to eliminate those.

Even then, it shouldn't be marked wrong since the expression itself is correct.

ehild
Homework Helper
Even then, it shouldn't be marked wrong since the expression itself is correct.
SammyS was right. The initial and final expressions are not identical, as their ranges are different. The final one is not defined when cos(x)=0. The initial one is not defined when either sin(x) or cos(x) is zero.