I am Completely Clueless in Calculus

1. Sep 17, 2004

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?

2. Sep 17, 2004

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.

3. Sep 17, 2004

modmans2ndcoming

mmmm...I love Trig Identities.

4. Sep 17, 2004

PrudensOptimus

Greetings friend,

Here is a step by step solution for your inquiry:

ln |sin(x)| = ln | (1-cos2x) / 2 | (after applying neccessary 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.

5. Sep 17, 2004

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

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

6. Sep 19, 2004

jamimi13

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

7. Sep 19, 2004

Pyrrhus

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

8. Sep 19, 2004

Theelectricchild

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