Trigonononometric Identities...

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frauleinedoctor

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I just need help knowing what to do first with many of these. I can figure it out usually once i get the first step or so, but these things take soooo long : |. Anybody got any suggestions?? heres one:

Factor and simplify (using the identities of course)

csc^4x-cot^4x
 


When we want to simplify a trigonometric expression that contains secant(s), cosecant(s), and/or cotangent(s), then a good first start is to rewrite the expression with everything in terms of sine and cosine.

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

Factor \(\displaystyle \frac{1}{sin^4(x)}\) from this expression.

The other factor will be a difference of squares, so factor it, too.

Finish by using the identity that states the sum of sine squared and cosine squared is 1 to do a cancellation.

 
frauleinedoctor said:
… when you said factor, what did you mean?
Click to expand...


\(\displaystyle \frac{1}{sin^4(x)} \cdot (A - B) = \frac{1}{sin^4(x)} \; - \; \frac{cos^4(x)}{sin^4(x)}\)

Factoring means finding expressions for A and B such that subsequent use of the distributive property (to multiply the two factors) returns to the unfactored form.

EG:

Factor \(\displaystyle \frac{cos(x)}{sin(x)}\) from \(\displaystyle \frac{cos(x)}{sin(x)} \; - \; \frac{cos^2(x)}{sin(x)}\)

The factorization is \(\displaystyle \frac{cos(x)}{sin(x)} \cdot [1 - cos(x)]\)

Do you see why?

 
What about cos^2xsin^2x - cos^2x ? How do you know what to even do first, there's so many things you can do.
 


Good.

Now do that factorization on your exercise.

The (A - B) part winds up being a difference of squares.

Do that factorization, too.

Then, you can factor \(\displaystyle \frac{1}{sin^4(x)}\) as \(\displaystyle \frac{1}{sin^2(x)} \cdot \frac{1}{sin^2(x)}\), followed by making use of the basic identity that I previously mentioned.

See the cancellation?

 
frauleinedoctor said:
umm...would it look like this? :

sin^4x ( 1 - cos^4x) ?
Click to expand...


No.

The factored form will look like the following.

\(\displaystyle \frac{1}{sin^4(x)} \cdot (A - B)\)

where

\(\displaystyle \frac{1}{sin^4(x)} \cdot A \; = \; \frac{1}{sin^4(x)}\)

and

\(\displaystyle \frac{1}{sin^4(x)} \cdot B \; = \; \frac{cos^4(x)}{sin^4(x)}\)

When you factor an expression, it's a good idea to (at least mentally) reverse your factorization by multiplying it out in order to verify that your factorization is correct.

If I do this with your factorization, I get the following (which is clearly incorrect).

\(\displaystyle sin^4(x) \cdot [1 - cos^4(x)] \; = \; sin^4(x) - sin^4(x) \cdot cos^4(x)\)

 
wait, so is it:

1/sin^4x (1 - cos^4x) ?

Because 1/sin^4x times 1 is just the same thing, and then 1/sin^4x times - cos^4x would be the - cos^4x/sin^4x that I started with, right?
 
frauleinedoctor said:
Because I took out the sin^4x from (1/sin^4x - cos^4x/sin^4x)
Click to expand...


We are not allowed to factor out the denominator only! You need to factor out the entire fraction \(\displaystyle \frac{1}{sin^4(x)}\)

What you did is like the following.

Factor 1/4 out of (1/4 - 3/7).

4(1 - 3/7)

Do you understand why this factorization does not equal (1/4 - 3/7)?

It equals 4 - 3/7, instead.

Please try again.

 
frauleinedoctor said:
… then I have to split 1/sin^4x into two parts to make it work?
Click to expand...


Yes.

I do it in my head, but I think that you should write out each step, for now.

 
frauleinedoctor said:
… so is it:

1/sin^4x (1 - cos^4x) ?
Click to expand...


By Jove, now I really think you've got it!

Do you see that 1 - cos^4(x) is a difference of squares?

Factor it.

 
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