Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is the problem: $$f(x)=\ln[\sin(-2x)\cos(-2x)]$$

This is as far as I can get: $$\frac{-2[\cos(-2x)]}{\sin(-2x)}+\frac{2[\sin(-2x)]}{\cos(-2x)}$$

I'm familiar with the rules of differentiation and the rules that apply to natural logs when we are to expand an expression. However, I'm at a point right now that I can't seem to take off in the right direction. I keep going off in the wrong direction, because I eventually end up with 0 over something, which I know is wrong. I know that I can simplify this further though. I also know, according to Wolfram Alpha that I will have $-4 \cot (4x)$ as my final answer. I see how $\cot$ can be derived from this, but I just can't get to it correctly. Can someone help me out.

Furthermore, I do not understand how I'm suppose to get the $4x$. For example I thought $$\sin(2x)\sin(2x) = \sin^2(2x)$$ and not $\sin(4x)$. This is just an example of what I'm talking about as it isn't in the actually problem below. However during the simplification something similar like this will have to take place so I assume.

In conclusion, could someone help me finish simplifying this expression and during the process explain how we end up with the $4x$ inside the cot function and when it's legal to multiply these terms inside the trig functions themselves, like I just talked about above.

NOTE: There is no need to go in depth to explain any trig identities or differentiation rules as I have a sheet with all of them directly in front of me.

share|improve this question
1  
Are you trying to differentiate? if so, the sin(-2x) *cos(-2x) = (1/2)sin(-4x) => double angle trig formula. after that, try it yourself it's pretty straightforward. you'll get -4cos(-4x)/sin(-4x) –  phoenix Apr 8 '13 at 1:59
1  
You are correct that $$\sin(2x)\sin(2x)=\sin^2(2x)$$ (that's literally the definition), and that this is not the same as $\sin(4x)$. The reason $4x$ appears is indicated in my hint below. –  Zev Chonoles Apr 8 '13 at 2:00
    
is this hint another identity. The only identity that I have on my sheet in front of me that resembles this is sin2x = 2sinxcosx –  Shane Yost Apr 8 '13 at 2:03
1  
@ShaneYost ^that one will work! –  phoenix Apr 8 '13 at 2:06
    
more help: after using the hint you'll get ln([sin(-4x)]/2) which by using the chain rule and differentiation, will be equal to 1/f(x) * f'(x) where f(x) is the function inside the natural log. ln(f(x)) = 1/f(x) * f'(x) –  phoenix Apr 8 '13 at 2:13
show 1 more comment

1 Answer

up vote 1 down vote accepted

Hint: $$\sin(2x)\cos(2x)=\frac{1}{2}\sin(4x)$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.