Little messy.
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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.Quote:
Originally Posted by Krizalid
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
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.Quote:
Originally Posted by Krizalid
did you have major problems decoding what i had?
A little bit, but try to convert it into MathType :D
ok, i'll look it up and tryQuote:
Originally Posted by Krizalid
let's see if this works:Quote:
Originally Posted by 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
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.
The idea, it's only handle the right side of the identity to get the another 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?Quote:
Originally Posted by Krizalid
Of course it is.Quote:
Originally Posted by Jhevon
:D
I forgot the explicit instruction :D
Nothing wrong with what I did.Quote:
Originally Posted by Krizalid
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 painQuote:
Originally Posted by ThePerfectHacker