Trigonometry - log/ln and absolute sign in equations

Question

Will this equation still hold if the absolute sign is being used at different places

For example,

This trigonometry identity;

$\frac{-1}{3}log|\frac{cos3x+1}{cos3x-1}|=\frac{2}{3}ln|(\frac{sin\frac{3x}{2}}{\cos\frac{3x}{2}})|$.

My question is

Does $\frac{-1}{3}log\frac{|cos3x|+1}{|cos3x-1|}=\frac{2}{3}ln(\frac{sin\frac{3x}{2}}{\cos\frac{3x}{2}})$?

Or even $\frac{-1}{3}log\frac{cos3x+1}{|cos3x-1|}=\frac{2}{3}ln(\frac{sin\frac{3x}{2}}{\cos\frac{3x}{2}})$?

Quite confused. Thank you in advance :)

Answer 1

Definitely not, you can't change the absolute value's position randomly, it's about as logical as changing a square's position randomly.

Graphical proof that your equalities do not hold :

Answer 2

$$1+cos(x)=2cos^(\frac{x}{2})\\1+cos(x)=2sin^2(\frac{x}{2})$$ so $$\frac{|cos(3x)+1|}{|cos(3x)-1|} =|\frac{cos(3x)+1}{cos(3x)-1}|=\frac{|2cos^2(\frac{3x}{2})|}{|-2sin^2(\frac{3x}{2})|}$$ as we know $|ab|=|a||b| \rightarrow |-a|=|-1||a|=|a|$ $$\frac{|2cos^2(\frac{3x}{2})|}{|-2sin^2(\frac{3x}{2})|}=\frac{|2cos^2(\frac{3x}{2})|}{|+2sin^2(\frac{3x}{2})|}=\\\frac{|cos^2(\frac{3x}{2})|}{|sin^2(\frac{3x}{2})|}=\frac{cos^2(\frac{3x}{2})}{sin^2(\frac{3x}{2})}=\\(\frac{cos(\frac{3x}{2})}{sin(\frac{3x}{2})})^2$$and now $$\log (\frac{cos(\frac{3x}{2})}{sin(\frac{3x}{2})})^2=\log(|\frac{cos(\frac{3x}{2})}{sin(\frac{3x}{2})}|)^2=2log(|\frac{cos(\frac{3x}{2})}{sin(\frac{3x}{2})}|)=-2\log (|\frac{sin(\frac{3x}{2})}{cos(\frac{3x}{2})}|)$$

Answer 3

The short answer is, no, the examples you gave are not valid identities.

Start with the trigonometric identity

$$-\frac13 \ln\left|\frac{\cos(3x)+1}{\cos(3x)-1}\right| =\frac23 \ln\left|\frac{\sin\frac{3x}{2}}{\cos\frac{3x}{2}}\right|.$$

In general, $\left|\frac ab\right| = \frac{|a|}{|b|},$ so you can write

$$-\frac13 \ln\frac{|\cos(3x)+1|}{|\cos(3x)-1|} =\frac23 \ln\frac{|\sin\frac{3x}{2}|}{|\cos\frac{3x}{2}|}.$$

Assuming we're working with real values only, $\cos(3x)+1 \geq 0$ always, so you can even write

$$-\frac13 \ln\frac{\cos(3x)+1}{|\cos(3x)-1|} =\frac23 \ln\frac{|\sin\frac{3x}{2}|}{|\cos\frac{3x}{2}|}. \tag 1$$

But that's just about all you can do in regard to changing the absolute value signs. Any of the remaining quantities inside absolute value signs in Equation (1) above might be negative, so removing the absolute value around that quantity might cause you to take the logarithm of a negative number, which is undefined. If you remove both absolute values on the right side of Equation (1) and just one of the two trigonometric functions has a negative value, again you will be taking the logarithm of a negative number.

As for replacing $|\cos(3x)+1|$ with $|\cos(3x)|+1$, in general, unless you know that $a$ and $b$ are both positive or both negative, you cannot know that $|a+b| = |a| + |b|$. If $\cos(3x) < 0$ (which can occur for some $x$) then replacing $|\cos(3x)+1|$ by $|\cos(3x)|+1$ will cause you to take the logarithm of a different value on the left side of the equation, whereas the value on the right side has not changed; since the equation was true before, it must now be false whenever $\cos(3x) < 0$, and it is not an identity.