Prove the following identity algebraically

[General_Sax]

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.

 

[rock.freak667]

try using $sin^2x+cos^2x=1$ and substitute for cos²x

 

[symbolipoint]

Another approach, from the lefthand side of the original given equation, do you see that
(csc²x-cot²x)²=the expression originally on the leftside.

 

[symbolipoint]

Yes, again, do as in #3. Then do the substitutions for the meaning of csc and for cot, simplify using algebra steps, and use the identity suggested in post #2. This becomes very uncomplicated.

 

[General_Sax]

$$\frac{2 - cos^2x + 2cos^4x}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}$$

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 $$- cos^2x$$ 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.

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

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

$$\frac{2 + cos^2x - (2sin^2xcos^2x)}{2sin^4x} = \frac{1 - cos^2x}{sin^2x}$$

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

I'm left with this:

$$\frac{2 + cos^2x - cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

which simplifies to:

$$\frac{2}{sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

Where did I go wrong?

 

[General_Sax]

Let me restart from here:

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

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

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

$$\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

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

$$\frac{2 - 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

$$\frac{2(1-cos^2x)}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

$$\frac{1-cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

Is this correct? Thanks again for the help guys.

 

[General_Sax]

Let me restart from here:

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

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

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

$$\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

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

$$\frac{2 - 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

$$\frac{2(1-cos^2x)}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

$$\frac{1-cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

Is this correct? Thanks again for the help guys.

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

$$\frac{2 - 3cos^2x + 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

$$\frac{2 - 2cos^2x}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

$$\frac{2(1-cos^2x)}{2sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

$$\frac{1-cos^2x}{sin^2x} = \frac{1 - cos^2x}{sin^2x}$$

 

[Chaos2009]

I would like to point out an error you made very early on.

You started with:
$$csc^{4}x - 2*csc^{2}x*cot^{2}x + cot^{4}x = 1$$

And somehow got to:
$$\frac{1}{sin^{4}x} - \frac{cos^{2}x}{2*sin^4{x}} + \frac{cos^{4}x}{sin^{4}x} = 1$$

How did your coefficient of the second term flip from the numerator to the denominator? It should read:
$$\frac{1}{sin^{4}x} - \frac{2*cos^{2}x}{sin^4{x}} + \frac{cos^{4}x}{sin^{4}x} = 1$$

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

 

[General_Sax]

Thanks for the constructive criticism.