any math people here???
Saturday, November 16, 2002 by Lord Vimal | Discussion: WinCustomize Talk
[admins, please! dont delete this thread!
]anyone can help me in proving this trig identity?
sec x + 1 - tan x 1 + cos x
----------------- = ---------
tan x - sec x + 1 sin x
if u have the time....
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WOM
Reply #2 Saturday, November 16, 2002 10:25 AM
Reply #2 Saturday, November 16, 2002 10:25 AM
That was 40 years ago. I think those brain cells have died! Don't remember what (sec) means. 
WOM
Reply #4 Saturday, November 16, 2002 10:29 AM
Reply #4 Saturday, November 16, 2002 10:29 AM
Thats right werewolf, I said some of those brain cells died off. 
WOM
Reply #5 Saturday, November 16, 2002 10:33 AM
Reply #5 Saturday, November 16, 2002 10:33 AM
I remember how to do the other functions, but forgot how to do secant. Oh well, it is early in the morning to do brain teasers.
UBoB
Reply #7 Saturday, November 16, 2002 10:51 AM
Reply #7 Saturday, November 16, 2002 10:51 AM
Here's what I can remember from 25 years ago:
I don't know how much you know so if anything below doesn't make sense then please write back.
A very effective way to prove most trig identities like the one above is to use complex numbers. To do this we use the formula
eix = cos(x) + isin(x)
So cos(x) = (1/2)(eix + e-ix)
and sin(x) = (1/2i)(eix - e-ix)
So if I set x = ip/11, then the question above is to show that:
(x4 - x-4)
------- + 2 (x1 - x-1) = i sqrt(11)
(x4 + x-4)
Now if we multiply both sides by (x4 + x-4) and then square each side and simplify then we get (after a little work which I'm certainly not typing out!):
4(x10 + x9 + x8 - x7 + x6 + 2x2 + 1 + x-10 + x-9 + x-8 - x-7 + x-6 + 2x-2) = 0
Now x11 = -1 (because x11 = eip = -1)
So -1 = x11 = x9+2 = x9 × x2 => x9 + x2 = 0
And similarly -x7 = x-4
So the long equation above becomes:
(x10 + x8 + x6 + x4 + x2 + x-10 + x-8 + x-6 + x-4 + x-2 + 1) = 0
So if we can prove this is true then the result you want will be true.
Now, x11 = -1 means that x22 = 1
So 0 = x22 - 1 = (x-1)(1 + x + x2 + x3 + ... + x21)
= (x-1)(1+x)(1 + x2 + x4 + x6 + ... + x20)
Now, clearly x isn't 1 or -1, so the above means that:
1 + x2 + x4 + ... + x20 = 0
Dividing by x10 (allowed as x isn't 0) gives:
x10 + x8 + x6 + x4 + x2 + x-10 + x-8 + x-6 + x-4 + x-2 + 1 = 0
Which is exactly what we wanted.
So tan(4p/11) + 4sin(p/11) = sqrt(11). QED
Whew!
I don't know how much you know so if anything below doesn't make sense then please write back.
A very effective way to prove most trig identities like the one above is to use complex numbers. To do this we use the formula
eix = cos(x) + isin(x)
So cos(x) = (1/2)(eix + e-ix)
and sin(x) = (1/2i)(eix - e-ix)
So if I set x = ip/11, then the question above is to show that:
(x4 - x-4)
------- + 2 (x1 - x-1) = i sqrt(11)
(x4 + x-4)
Now if we multiply both sides by (x4 + x-4) and then square each side and simplify then we get (after a little work which I'm certainly not typing out!):
4(x10 + x9 + x8 - x7 + x6 + 2x2 + 1 + x-10 + x-9 + x-8 - x-7 + x-6 + 2x-2) = 0
Now x11 = -1 (because x11 = eip = -1)
So -1 = x11 = x9+2 = x9 × x2 => x9 + x2 = 0
And similarly -x7 = x-4
So the long equation above becomes:
(x10 + x8 + x6 + x4 + x2 + x-10 + x-8 + x-6 + x-4 + x-2 + 1) = 0
So if we can prove this is true then the result you want will be true.
Now, x11 = -1 means that x22 = 1
So 0 = x22 - 1 = (x-1)(1 + x + x2 + x3 + ... + x21)
= (x-1)(1+x)(1 + x2 + x4 + x6 + ... + x20)
Now, clearly x isn't 1 or -1, so the above means that:
1 + x2 + x4 + ... + x20 = 0
Dividing by x10 (allowed as x isn't 0) gives:
x10 + x8 + x6 + x4 + x2 + x-10 + x-8 + x-6 + x-4 + x-2 + 1 = 0
Which is exactly what we wanted.
So tan(4p/11) + 4sin(p/11) = sqrt(11). QED
Whew!
UBoB
Reply #8 Saturday, November 16, 2002 10:55 AM
Reply #8 Saturday, November 16, 2002 10:55 AM
BTW this was an example of how to prove tan(4p/11) + 4sin(p/11) = sqrt(11). But it should show you how to go about proving yours. I was always taught by my trig teacher to "do your own work". 
Fuzzy Logic
Reply #10 Saturday, November 16, 2002 11:04 AM
Reply #10 Saturday, November 16, 2002 11:04 AM
I think I speak for everyone here when I say HUH?
Fuzzy Logic
Reply #11 Saturday, November 16, 2002 11:04 AM
Reply #11 Saturday, November 16, 2002 11:04 AM
I'm not very bright but I can lift heavy things...
Lord Vimal
Reply #12 Saturday, November 16, 2002 2:02 PM
Reply #12 Saturday, November 16, 2002 2:02 PM
hmm
thanks motion but i we havent learnt complex numbers yet
maybe i should consult with my teacher again!
thanks motion but i we havent learnt complex numbers yet
maybe i should consult with my teacher again!
Lord Vimal
Reply #13 Saturday, November 16, 2002 2:04 PM
Reply #13 Saturday, November 16, 2002 2:04 PM
actually the text is 'bad'. this is the identity:
sec(x) + 1 - tan(x)
------------------- equal to:
tan(x) - sec(x) + 1
=
1 + cos(x)
----------
sin(x)
sec(x) + 1 - tan(x)
------------------- equal to:
tan(x) - sec(x) + 1
=
1 + cos(x)
----------
sin(x)
Lord Vimal
Reply #14 Saturday, November 16, 2002 2:06 PM
Reply #14 Saturday, November 16, 2002 2:06 PM
we have to prove this using these identities (we have learnt)
sin²x + cos²x = 1
1 + tan²x = sec²x
1 + cot²x = cosec²x
but still not gettin it! >
sin²x + cos²x = 1
1 + tan²x = sec²x
1 + cot²x = cosec²x
but still not gettin it! >
Lord Vimal
Reply #15 Saturday, November 16, 2002 2:23 PM
Reply #15 Saturday, November 16, 2002 2:23 PM
yay!!!!! i had actually asked a math doctor on the internet. he just guided me. i found it out. here's it:
1+cos(x)-sin(x)
---------------
sin(x)-1+cos(x)
= (1+cos(x)) - sin(x) || (1+cos(x))+sin(x)
------------------- || -----------------
sin(x)-(1-cos(x)) || (1+cos(x))+sin(x)
= 1+cos²(x)+2cos(x)-1+cos²(x)
---------------------------
sin²(x)+sin(x)[1+cos(x)]-sin²(x)-sin(x)[1-cos(x)]
= 2cos(x)[1+cos(x)]
-----------------
2sin(x).cos(x)
= 1+cos(x)
--------
sin(x)
= RHS!! yay! lol
1+cos(x)-sin(x)
---------------
sin(x)-1+cos(x)
= (1+cos(x)) - sin(x) || (1+cos(x))+sin(x)
------------------- || -----------------
sin(x)-(1-cos(x)) || (1+cos(x))+sin(x)
= 1+cos²(x)+2cos(x)-1+cos²(x)
---------------------------
sin²(x)+sin(x)[1+cos(x)]-sin²(x)-sin(x)[1-cos(x)]
= 2cos(x)[1+cos(x)]
-----------------
2sin(x).cos(x)
= 1+cos(x)
--------
sin(x)
= RHS!! yay! lol
Lord Vimal
Reply #16 Saturday, November 16, 2002 2:24 PM
Reply #16 Saturday, November 16, 2002 2:24 PM
Lord Vimal got it!!
problem with the text >
problem with the text >
Jafo
Reply #18 Saturday, November 16, 2002 9:24 PM
Reply #18 Saturday, November 16, 2002 9:24 PM
31 years ago I may have known this one....but if you don't use it, it drops off...[he says in a high, squeaky voice]...
Oh, yeah....it's nearly always '42'...
Oh, yeah....it's nearly always '42'...
Lord Vimal
Reply #19 Saturday, November 16, 2002 11:17 PM
Reply #19 Saturday, November 16, 2002 11:17 PM
hey whats the '42' being talked about here...?
Lockie56
Reply #20 Sunday, November 17, 2002 12:33 AM
Reply #20 Sunday, November 17, 2002 12:33 AM
It's the answer to "What is the meaning of Life?"
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Reply #1 Saturday, November 16, 2002 10:16 AM
Doreen only goes as far as 2 + 2 = 3