# Interesting trig. proof

• Mar 31st 2007, 09:10 PM
Krizalid
Interesting trig. proof
Little messy.
• Mar 31st 2007, 11:24 PM
Jhevon
Quote:

Originally Posted by Krizalid
Little messy.

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, 07:13 AM
Krizalid
Yeah, but you can use MathType while LaTeX comes back.

It's for better understanding.
• Apr 1st 2007, 07:15 AM
Jhevon
Quote:

Originally Posted by Krizalid
Yeah, but you can use MathType while LaTeX comes back.

It's for better understanding.

Yeah, i guess, i never learnt how to use any of those "languages" though. i guess i'll check it out and see.

did you have major problems decoding what i had?
• Apr 1st 2007, 07:18 AM
Krizalid
A little bit, but try to convert it into MathType :D
• Apr 1st 2007, 07:24 AM
Jhevon
Quote:

Originally Posted by Krizalid
A little bit, but try to convert it into MathType :D

ok, i'll look it up and try
• Apr 1st 2007, 07:57 AM
Jhevon
Quote:

Originally Posted by Jhevon
ok, i'll look it up and try

let's see if this works:
• Apr 1st 2007, 07:57 AM
Jhevon
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, 08:12 AM
ThePerfectHacker
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, 08:29 AM
Krizalid
The idea, it's only handle the right side of the identity to get the another one.
• Apr 1st 2007, 08:31 AM
Jhevon
Quote:

Originally Posted by Krizalid
The idea, it's only handle the right side of the identity to get the other one.

working on both sides to get them the same is a valid proof technique. was working on the right side only an explicit instruction?
• Apr 1st 2007, 08:35 AM
Krizalid
Quote:

Originally Posted by Jhevon
working on both sides to get them the same is a valid proof technique.

Of course it is.

:D

I forgot the explicit instruction :D
• Apr 1st 2007, 08:35 AM
ThePerfectHacker
Quote:

Originally Posted by Krizalid
The idea, it's only handle the right side of the identity to get the another one.

Nothing wrong with what I did.

What would have been wrong is two square both sides.
(Unless I can show they always have the same sign, which is going to be a bigger mess).
• Apr 1st 2007, 08:36 AM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
Nothing wrong with what I did.

What would have been wrong is two square both sides.
(Unless I can show they always have the same sign, which is going to be a bigger mess).

no one said what you did was wrong, he just wanted to get an asnwer using a different route, which as far as i can tell is going to be a real pain