1. ## Interesting trig. proof

Little messy.

2. 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

3. Yeah, but you can use MathType while LaTeX comes back.

It's for better understanding.

4. 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?

5. A little bit, but try to convert it into MathType

6. Originally Posted by Krizalid
A little bit, but try to convert it into MathType
ok, i'll look it up and try

7. Originally Posted by Jhevon
ok, i'll look it up and try
let's see if this works:

8. It's a little small, but I think you can read it, right?

Hey, i'm a first-timer man, cut me some slack

9. 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.

10. The idea, it's only handle the right side of the identity to get the another one.

11. 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?

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

I forgot the explicit instruction

13. 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).

14. 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