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## Homework Statement

Must be proven algebraically, duh!

## Homework Equations

trig identities

## The Attempt at a Solution

I'm at a loss as what to do next. Any help would be appreciated.

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- Thread starter General_Sax
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- #1

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Must be proven algebraically, duh!

trig identities

I'm at a loss as what to do next. Any help would be appreciated.

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- #2

rock.freak667

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try using [itex]sin^2x+cos^2x=1[/itex] and substitute for cos^{2}x

- #3

symbolipoint

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Another approach, from the lefthand side of the original given equation, do you see that

(csc^{2}x-cot^{2}x)^{2}=the expression originally on the leftside.

(csc

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- #4

symbolipoint

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- #5

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How am I going to get rid of the two on the L.H.S.? I can't just directly cancel it, because of the [tex] - cos^2x [/tex] right?

I'm at a loss, even with all this help. So I'm going to show you my next step and maybe you people can tell me what I'm doing wrong.

[tex] \frac{2 - [cos^2x ( 1 + 2cos^2x )]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [cos^2x ( 1 + 2(1 - sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 + cos^2x - (2sin^2xcos^2x)}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

I do this next: I cancel the two coefficients in the denominator and in the bracketed term, then I subtract the exponent from the [tex]sin^2x[/tex] in the brackets from the denominator.

I'm left with this:

[tex] \frac{2 + cos^2x - cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

which simplifies to:

[tex] \frac{2}{sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

Where did I go wrong?

- #6

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Let me restart from here:

[tex] \frac{2 - [cos^2x ( 1 + 2(1 - sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [cos^2x ( 1 + 2 - 2sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [3cos^2x - 2sin^2xcos^2x]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

In the previous step I subtracted the exponents for sin in the denominator from the numerator. Is this a legal maneuver?

[tex]\frac{2 - 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{2(1-cos^2x)}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{1-cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

Is this correct? Thanks again for the help guys.

[tex] \frac{2 - [cos^2x ( 1 + 2(1 - sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [cos^2x ( 1 + 2 - 2sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [3cos^2x - 2sin^2xcos^2x]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

In the previous step I subtracted the exponents for sin in the denominator from the numerator. Is this a legal maneuver?

[tex]\frac{2 - 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{2(1-cos^2x)}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{1-cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

Is this correct? Thanks again for the help guys.

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- #7

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Let me restart from here:

[tex] \frac{2 - [cos^2x ( 1 + 2(1 - sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [cos^2x ( 1 + 2 - 2sin^2x)]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex] \frac{2 - [3cos^2x - 2sin^2xcos^2x]}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

In the previous step I subtracted the exponents for sin in the denominator from the numerator. Is this a legal maneuver?

[tex]\frac{2 - 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{2(1-cos^2x)}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{1-cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

Is this correct? Thanks again for the help guys.

I made in error in writing down my work, it should read:

[tex]\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{2 - 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{2(1-cos^2x)}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

[tex]\frac{1-cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}[/tex]

- #8

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You started with:

[tex]csc^{4}x - 2*csc^{2}x*cot^{2}x + cot^{4}x = 1[/tex]

And somehow got to:

[tex]\frac{1}{sin^{4}x} - \frac{cos^{2}x}{2*sin^4{x}} + \frac{cos^{4}x}{sin^{4}x} = 1[/tex]

How did your coefficient of the second term flip from the numerator to the denominator? It should read:

[tex]\frac{1}{sin^{4}x} - \frac{2*cos^{2}x}{sin^4{x}} + \frac{cos^{4}x}{sin^{4}x} = 1[/tex]

From there it is pretty straightforward. Add the terms together, factor the top, then use the Pythagorean trig identities to reduce it to 1.

- #9

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Thanks for the constructive criticism.

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