Trigonononometric Identities...

Haha ok. So now, I split up 1/sin^4x into two parts. Should I also split up (1-cos^4x) into (1 + cos^2x) (1-cos^2x) ?
 
frauleinedoctor said:
… I split up 1/sin^4x into two parts. Should I also split up (1-cos^4x) into (1 + cos^2x) (1-cos^2x) ?
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Yes. That's the factorization of the difference of squares.

Since you factored the fraction with sine, also, you can now write the product as follows.

\(\displaystyle \frac{1 + cos^2(x)}{sin^2(x)} \cdot \frac{1 - cos^2(x)}{sin^2(x)}\)

One last step, and your original exercise is simplified.

After I see your final result, I'm going to show you another way that's a bit simpler.

 
frauleinedoctor said:
… how did you get that?
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I rearranged the factors using the Commutative Property of Multiplication.

1/y^2 * (1 - x) * (1 + x)

(1/y) * (1/y) * (1 - x) * (1 + x)

(1/y) * (1 - x) * (1/y) * (1 + x)

(1 - x)/y * (1 + x)/y

 
frauleinedoctor said:
… wait .nevermind I see why.
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I'm too fast to nevermind.

(Me thinks that I should have started with the simpler way.)

So, do you see the cancellation now?

 
frauleinedoctor said:
Ok, I'm at

1 + cos^2x / 1 - cos^2x and feeling a little stuck.
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Yikes! :eek:

Okay, at this point, I think that I should finish this exercise for you.

We left off with this:

\(\displaystyle \frac{1 + cos^2(x)}{sin^2(x)} \cdot \frac{1 - cos^2(x)}{sin^2(x)}\)

Now, here is a basic identity.

\(\displaystyle sin^2(x) + cos^2(x) = 1\)

From this, we get the following two identities.

\(\displaystyle sin^2(x) = 1 - cos^2(x)\)

\(\displaystyle cos^2(x) = 1 - sin^2(x)\)

From the former of the two, we see that the fraction \(\displaystyle \frac{1 - cos^2(x)}{sin^2(x)}\) must equal 1.

Therefore, the answer to your exercise is:

\(\displaystyle \frac{1 + cos^2(x)}{sin^2(x)}\)

 
I am so never going to finish my hw : ( My brain is missing the attachment designed for this stuf : {
 


Here's another way that may seem simpler, to you.

\(\displaystyle \frac{1}{sin^4(x)} - \frac{cos^4(x)}{sin^4(x)}\)

Since we have a common denominator, we can go ahead and do the subtraction.

\(\displaystyle \frac{1 - cos^4(x)}{sin^4(x)}\)

Now factor the numerator and the denominator.

\(\displaystyle \frac{[1 + cos^2(x)] \cdot [1 - cos^2(x)]}{sin^2(x) \cdot sin^2(x)}\)

The cancellation finishes it.

\(\displaystyle \frac{1 + cos^2(x)}{sin^2(x)}\)

These steps are basically the same without the initial factorization of 1/sin^4(x). I combined the two fractions to start, instead.

 
frauleinedoctor said:
Your going to seriously have to walk me through that one, man.
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I'm willing to do that, but you're going to seriously have to start asking specific questions so that I can tell what it is that you do not understand.

Otherwise, I will tell you to reread the entire thread. Everything is here for you to see.

What is it that you do not understand?

 
Seriously? Thats it?? Wow. Indeed, that IS simpler. But the answer you came to isn't any of the answers given on the homework. Its multiple choice, and the answers given are

a) cot^2x

b ) csc^2xcos^2x

c) csc^2x + cot^2x and

d) 1
 
Wait a min, isn't 1 + cos^2 / 1 - cos^2 the same thing as the final answer you gave? Yours is just simplified more, right? So i wasn't alll that far off, right?
 
frauleinedoctor said:
… But the answer you came to isn't any of the answers given on the homework …
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Yes, it is.

We can decompose our result back into a difference of two fractions.

\(\displaystyle \frac{1 + cos^2(x)}{sin^2(x)} = \frac{1}{sin^2(x)} + \frac{cos^2(x)}{sin^2(x)}\)

Now rewrite these two terms using the definitions for cosecant and cotangent. (Sort of the reverse of what we did at the very beginning with the original expression.)

MY EDIT: Fixed typographical error (minus sign) in LaTex
 
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