Little messy.

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- Mar 31st 2007, 10:10 PMKrizalidInteresting trig. proof
Little messy.

- Apr 1st 2007, 12:24 AMJhevon
Forgive me Krizalid, but as you know, we can't use LaTex, it might take you a while to decode what I did.

the question is:

Prove: (sin(x) - cos(x))/sqrt(sin(2x)) = [sin(x) - cos(x) - cot(2x)*sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))]

Consider RHS:

[sin(x) - cos(x) - cot(2x)*sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))]

= [sin(x) - cos(x) - {cos(2x)/sin(2x)}*sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))]

= [sin(x) - cos(x) - cos(2x)/sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))]

= [{sin(x)*sqrt(sin(2x)) - cos(x)*sqrt(sin(2x)) - cos(2x)}/sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))] .......combine the top fractions

= [sin(x)*sqrt(sin(2x)) - cos(x)*sqrt(sin(2x)) - cos(2x)]/[{sin(x) + cos(x) + sqrt(sin(2x))}*sqrt(sin(2x))] .......combine the top and bottom into one fraction

Let's leave that one there

Now, consider the LHS:

(sin(x) - cos(x))/sqrt(sin(2x))

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

= [sin^2(x) + sin(x)cos(x) + sin(x)*sqrt(sin2x)) - sin(x)cos(x) - cos^2(x) - cos(x)*sqrt(sin(2x))]/[{sin(x) + cos(x) + sqrt(sin(2x))}*sqrt(sin(2x))]

= [-(cos^2(x) - sin^2(x)) + sin(x)*sqrt(sin2x)) - cos(x)*sqrt(sin(2x))]/[{sin(x) + cos(x) + sqrt(sin(2x))}*sqrt(sin(2x))]

= [-cos(2x) + sin(x)*sqrt(sin2x)) - cos(x)*sqrt(sin(2x))]/[{sin(x) + cos(x) + sqrt(sin(2x))}*sqrt(sin(2x))]

now we see that both sides "simplify" to the same thing. thus LHS = RHS

QED - Apr 1st 2007, 08:13 AMKrizalid
Yeah, but you can use MathType while LaTeX comes back.

It's for better understanding. - Apr 1st 2007, 08:15 AMJhevon
- Apr 1st 2007, 08:18 AMKrizalid
A little bit, but try to convert it into MathType :D

- Apr 1st 2007, 08:24 AMJhevon
- Apr 1st 2007, 08:57 AMJhevon
- Apr 1st 2007, 08:57 AMJhevon
It's a little small, but I think you can read it, right?

Hey, i'm a first-timer man, cut me some slack - Apr 1st 2007, 09:12 AMThePerfectHacker
Nothing complicated.

I did,

LHS*(denominator RHS)/(denominator RHS) = RHS*(denominator LHS)/(denominator LHS)

Of course the denominators now match up.

And will a little work the numerators do too. - Apr 1st 2007, 09:29 AMKrizalid
The idea, it's only handle the right side of the identity to get the another one.

- Apr 1st 2007, 09:31 AMJhevon
- Apr 1st 2007, 09:35 AMKrizalid
- Apr 1st 2007, 09:35 AMThePerfectHacker
- Apr 1st 2007, 09:36 AMJhevon