I am Completely Clueless in Calculus

[jamimi13]

Today i was given some review questions from pre-calc, but i forgot how to even start the problems. One problem looks like this:
ln lsinxl=(ln l1-cos2x)-ln2)


i know that i have to prove that each side is equal, but i don't know where to begin...any suggestions?

 

[Pyrrhus]

$$\ln |\sin(x)| = \ln |1-cos2x| - \ln(2)$$

Apply Logarithm properties, (Review them)

$$\ln |\sin(x)| = \ln |\frac{1-cos2x}{2}|$$

Here's another hint Look up Half-Angle identities.

 

[modmans2ndcoming]

$$\ln |\sin(x)| = \ln |1-cos2x| - \ln(2)$$

Apply Logarithm properties, (Review them)

$$\ln |\sin(x)| = \ln |\frac{1-cos2x}{2}|$$

Here's another hint Look up Half-Angle identities.

mmmm...I love Trig Identities.

 

[PrudensOptimus]

Greetings friend,

Here is a step by step solution for your inquiry:

ln |sin(x)| = ln | (1-cos2x) / 2 | (after applying necessary log laws)
you raise base e to some exponets:
e^(ln|sin(x)|) = e^(ln|(1-cos2x)/2)
sin(x) = (1-cos2x)/2 (notice 1-cos2x/2 is another way of saying sin^2(x))

so they are not equal... unless you made a type and meant ln|sin^2(x)| originally.

 

[relinquished™]

PrudensOptimus is correct. There must be something wrong with your equality, unless the question was if you were to prove or disprove it. If the equality should be true, then

$$<br /> ln |sin^2 x| = ln | 1 - cos2x | - ln (2)<br />$$

 

[jamimi13]

That's for all the help guys!
You are all lifesavers!

 

[Pyrrhus]

That's for all the help guys!
You are all lifesavers!

It's glad to be of help!, and Welcome to PF!, i hope you enjoy your stay

 

[Theelectricchild]

This is the second best site in the world, after google of course.