A pretty difficult trig identity

Originally posted by wolfgang59
What you have here is an equality which needs proving NOT an identity which as previously discussed will involve a variable.

eg x^2 - y^2 = (x-y)(x+y) is an identity


That of course doesnt make it any easier! 😞

My algebra book says it's an identity. Ok, in Latvian it sounds and spells very similar but, ok, English is not my language, so I'll assume it's an equality, not an identity.

Your proof, Mtthw, looks sound to me but the proof I intended is a bit different.

It can be proved with the use of these formulas (which we all have to know in school) by simply rearranging the expression a little:

cosx + cosy = 2 * cos((x+y)/2) * cos((x-y)/2)
sin2x = 2 * sinx * cosx
sinx + siny = 2 * sin((x+y)/2) * cos((x-y)/2)

EDIT: sinx * cosx = 1/2 * (sin(x + y) + sin(x - y))

I think that's all you need to know, to prove this equality.

Originally posted by mtthw
Having tried various different approaches - and proving various identities that weren't asked for (I'm seriously out of practice with these things) - I've got a method that's so simple I suspect there's something wrong with it!

[Will write 2pi/7 as a, to simplify things]

It relies on e^(ia) being a root of x^7 = 1.

But x^7 = (x - 1)(x^6 + x^5 ...[text shortened]... s(a) + cos(2a) + cos(3a)] + 1 = 0

=> cos(2pi/7) + cos(4pi/7) + cos(6pi/7) = -1/2
Ta-da!

yes u r perfectly correct in ur method, but u have to come up with e^(ia) is a root of x^7=1 originally out of somewhere..i think cos(3x)=cos(4x) when x=2n(pi)/7 comes to mind more naturally than that, dont u think..anyway congrats and thanks for this different method!
Cheers!

Originally posted by blacknight1985
yes u r perfectly correct in ur method, but u have to come up with e^(ia) is a root of x^7=1 originally out of somewhere..i think cos(3x)=cos(4x) when x=2n(pi)/7 comes to mind more naturally than that, dont u think..anyway congrats and thanks for this different method!
Cheers!

Looking for complex roots of unity is quite a common method in this sort of problem - a technique I'd forgotten until I'd tried a couple of other things, admittedly. Reading the thread again, I notice that Jirakon had already picked up on that.

Oh, and for what it's worth, it is an identity. It doesn't matter that there's no x in it. The LHS and RHS are equivalent.

Originally posted by mtthw
Oh, and for what it's worth, it is an identity. It doesn't matter that there's no x in it. The LHS and RHS are equivalent.

Finally someone agrees with me on this.


Now, can you prove it with the formulas I provided?

P.s. There, in that post, is an EDIT now.

Originally posted by mtthw
Oh, and for what it's worth, it is an identity. It doesn't matter that there's no x in it. The LHS and RHS are equivalent.

definition 6a from
http://www.answers.com/topic/identity?cat=technology

and from good ol' wiki:
"An identity is an equality that remains true regardless of the values of any variables that appear within it, to distinguish it from an equality which is true under more particular conditions"
(http://en.wikipedia.org/wiki/Identity_%28mathematics%29)

and finally
Math. an equation which is true for all permissible sets of values of the variables which appear in it: Ex.: x - y= (x + y) (x - y)
(http://www.yourdictionary.com/identity)

Originally posted by wolfgang59
and from good ol' wiki:
"An identity is an equality that remains true regardless of the values of any variables that appear within it, to distinguish it from an equality which is true under more particular conditions"
(http://en.wikipedia.org/wiki/Identity_%28mathematics%29)

I think you'll find this fits that definition. There are no variables, so it's definitely true regardless of their values. It's not an equality because there are no conditions under which it isn't true.

Originally posted by wolfgang59
definition 6a from
http://www.answers.com/topic/identity?cat=technology

and from good ol' wiki:
"An identity is an equality that remains true regardless of the values of any variables that appear within it, to distinguish it from an equality which is true under more particular conditions"
(http://en.wikipedia.org/wiki/Identity_%28mathematics%29)
...[text shortened]... les which appear in it: Ex.: x - y= (x + y) (x - y)
(http://www.yourdictionary.com/identity)

The bold text under your nick says it all.

Can you prove the equality/identity (whatever it is) without using a calculator?

EDIT: This is from the first link you provided:

* An identity is an equality that remains true regardless of the values of any variables that appear within it, to distinguish it from an equality which is true under more particular conditions. For this, the symbol ≡ (the congruence symbol us supposed to be there, '=' with a third horizontal line) is sometimes used. (However, this can be ambiguous since the same symbol can also be used for a congruence relation.)

Originally posted by kbaumen
Can you prove the equality/identity (whatever it is) without using a calculator?

Of course. I do most of my radian measure calculations in my head.

Originally posted by kbaumen
[b]The bold text under your nick says it all.

Yeh. ... I wonder who put that there?

Wait a little. Do I hear someone saying that you can use a calculator to prove any identity (like 2+2=4) if the expressions are in constants and not variables?

A calculator can't even prove that (1/3)*3=1!
A calculator cannot hold the exact value of 1/3 in its memory, only a approximation. Therefore, multiplying this approximate number with 3 will still be an approximate number. Doesn't care if you get 1 as the result, the calculator only round it to 1, it's not really 1 exactly.

So when we talk about the cos(2pi/7) + cos(4pi/7) + cos(6pi/7) = -1/2
identity (from the first posting of this thread), we cannot use a calculator at all. We can use a calculator to make it probable that it is an identity, but not as a proof.

Originally posted by FabianFnas
Wait a little. Do I hear someone saying that you can use a calculator to prove any identity (like 2+2=4) if the expressions are in constants and not variables?

A calculator can't even prove that (1/3)*3=1!
A calculator cannot hold the exact value of 1/3 in its memory, only a approximation. Therefore, multiplying this approximate number with 3 ...[text shortened]... all. We can use a calculator to make it probable that it is an identity, but not as a proof.

Very true. While we're being this precise, however, we should say that we cannot use a calculator in a finite amount of time to prove the identity.

In that vain, can anyone construct a finite-state automaton to prove this identity, or even to prove that 1/3*3 = 1?

Originally posted by ChronicLeaky
Very true. While we're being this precise, however, we should say that we cannot use a calculator in a finite amount of time to prove the identity.

In that vain, can anyone construct a finite-state automaton to prove this identity, or even to prove that 1/3*3 = 1?

As computers are using binary memories, 1/3 can't be expressed exactly.
However, a computer with trinary memories (where a bit can store one out of three states), 1/3 is perfectly possible to store (as 1/3 is 0.1 trinary).
But this trinary computer can't prove the (1/2)*2=1 identity as 1/2 cannot be exactly represented in trinary memories.

Originally posted by FabianFnas
As computers are using binary memories, 1/3 can't be expressed exactly.
However, a computer with trinary memories (where a bit can store one out of three states), 1/3 is perfectly possible to store (as 1/3 is 0.1 trinary).
But this trinary computer can't prove the (1/2)*2=1 identity as 1/2 cannot be exactly represented in trinary memories.

Well, you can buy calculators easily enough that can accurately do fractional arithmetic - they just don't store them that way. You can accurately store a fraction as two integers - numerator and denominator.

Originally posted by mtthw
Well, you can buy calculators easily enough that can accurately do fractional arithmetic - they just don't store them that way. You can accurately store a fraction as two integers - numerator and denominator.

Yea, I have a Citizen calculator that operates with fractions. No problem to, for example, add 1/2 + 1/3, it really gives 5/6. That's an easy algorithm to write.

In programming (Pascal) I had a homework to write a program in which the user has to enter the dividend and divisor and the result comes out as a fraction. Quite simple actually. So there should no problem writing a program that allows you to use other operations with fractions.

P.S.

I haven't still seen any decent proof involving the formulas I posted.