Two Identities

Hey guys, you didn't fail me last time, let's try this again.

1: cos(2X)=cos^4(X)-sin^4(X)
All I've really done on this one is that I factored the ___^4 things, but that didn't really help all that much. Could there possibly be a cos-sin formula involved?


2: csc(2X)-cot(2X) = tan(X)
For this, I did the (hopefully correct) next step of:
((2X)/sin(2x))-((cos(2X)/(sin(2X)) = tan(x)

then I tried this...
(2(x)-cos(2x))/(2sin(x)cos(x))

Is this correct? I'm stuck. please help!

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Originally Posted by evenkeel3x

cos(2X) = cos^4(x) - sin^4(x)

All I've really done on this one is that I factored the ___^4 things, but that didn't really help all that much.

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I think that you gave up too soon, but I'm not sure because you did not show your result after factoring.

Did you get the following?

cos(2x) = [cos^2(x) + sin^2(x)] * [cos^2(x) - sin^2(x)]

You should have because the given RHS is a difference of squares.

Now, look at this factorization carefully, in conjunction with the fundamental and double-angle identities in your textbook.

(Is there some significance to switching back and forth between the symbols x and X ?) :?

Quote:

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Originally Posted by evenkeel3x

Hey guys, you didn't fail me last time, let's try this again.

1: cos(2X)=cos^4(X)-sin^4(X)
All I've really done on this one is that I factored the ___^4 things, but that didn't really help all that much. Could there possibly be a cos-sin formula involved?


2: csc(2X)-cot(2X) = tan(X)
For this, I did the (hopefully correct) next step of:
((2X)/sin(2x))-((cos(2X)/(sin(2X)) = tan(x)

then I tried this...
(2(x)-cos(2x))/(2sin(x)cos(x))

Is this correct? I'm stuck. please help!

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You didn't show us how you factored the "____^4" things...please do. That should help immensely.

For problem #2, csc(2x) = 1 / sin(2x)....NOT 2X/(sin 2x) AND...X and x are different variables!

You could try this...work with the left side, and rewrite csc 2x - cot 2x in terms of sin 2x and cos 2x:

1/(sin 2x) - [cos 2x / sin 2x]

The denominators are the same, so combine the numerators as indicated, and put the result over the common denominator:

[1 - cos 2x) / (sin 2x)

Look at the double-angle identities. sin 2x = 2 sin x cos x, and cos 2x = 1 - sin^2 x. Substitute into the expression you have on the left side:

[1 - (1 - 2 sin^2 x)] / (2 sin x cos x)

Ok...now see if you can simplify that.

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Originally Posted by evenkeel3x

2: csc(2X)-cot(2X) = tan(X)

2X/sin(2x) - cos(2X)/sin(2X) = tan(x) I think that should be 1.

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I removed all of the unnecessary parentheses because it's easier to read typed expressions without them.

On the LHS, we have two ratios with a common denominator. Combine these into a single ratio.

[1 - cos(2x)]/sin(2x) = sin(x)/cos(x)

Use the double-angle indentities for cos(2x) and sin(2x). After that, it's an algebraic simplification.

Okay, so I suppose I'm really dumb or something, but which Double angle formula do you use for cos(2x) ? I've tried all three now and none of them are working. Help?

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Originally Posted by evenkeel3x

which Double angle formula do you use for cos(2x) ? I've tried all three

Use the identity for cos(2x) that allows you to make cancellations,
arriving at sin(x)/cos(x).

I mean, you have three factors in the denominator, yes?

2 * sin(x) * cos(x)

You want only cos(x) in the denominator, yes?

This means that you need to find some way to cancel both the factor of 2
and the factor of sin(x).

So, when you're looking at the possibilities for cos(2x), be thinking:

"I need a factor of 2 on top (to cancel), and
I need a factor of sin(x) on top (to cancel)".

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If you would like more help, please show us what you tried.

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"I need a factor of 2 on top (to cancel), and
I need a factor of sin(x) on top (to cancel)".

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if the cosine Double Angle Formulas are:
cos^2X-sin^2X
2cos^2X-1
1-2sin^2X

how does any of those give me a 2 or a sinX, in a form where i can cancel it?
They're all sutracted at the top, and I would need a monomial at the top to cancel out with the monomial at the bottom, am I correct?

Quote:

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Originally Posted by evenkeel3x

Quote:

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"I need a factor of 2 on top (to cancel), and
I need a factor of sin(x) on top (to cancel)".

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if the cosine Double Angle Formulas are:
cos^2X-sin^2X
2cos^2X-1
1-2sin^2X

how does any of those give me a 2 or a sinX, in a form where i can cancel it?
They're all sutracted at the top, and I would need a monomial at the top to cancel out with the monomial at the bottom, am I correct?

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As mmm4444bot suggested, you pick one that has "sin x" and "2" in it....and I went even further in my response....I SHOWED you which one to use, and where to put it, even.

All that was left for you to do was to simplify the numerator, as I suggested (by performing the subtraction you're so concerned about.) Did you DO that? Did you even READ my response? After simplifying the numerator you SHOULD come up with a monomial, which contains some of the same factors you see in the denominator.

Allright, thanks for the help on the second one, but do you have any advice on the first one after I get:
(cos^2x+sin^2x)(cosx-sinx)(cosx+sinx)

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Originally Posted by evenkeel3x

do you have any advice on the first one Yes, I do.

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It's the same advice that I gave you earlier. You should slow down and read the responses that people here have provided. I don't mean "skim" through the information; I mean study what you're reading. If you read something that you do not understand, then the appropriate response is to ask specific questions.

Here is my advice:

Quote:

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Originally Posted by mmm4444bot

cos(2x) = [cos^2(x) + sin^2(x)] * [cos^2(x) - sin^2(x)]

Now, look at this factorization carefully, in conjunction with the fundamental and double-angle identities in your textbook.

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In other words, compare each of the two factors above to the identities in your text. Do you "see" the two factors? Do you know what factors are? (Nobody suggested that you continue factoring.)

When you make no statements about why you're stuck, we can't possibly know why, either.