Help with Proof

[Eldin]

Any math whizzes out there? Could you help me prove that:

Cos (pi/5) = (1 + square root of 5)/4

Note that the second expression is the quantity of one plus the square root of five divided by four (hence the parenthesis).

Much appreciated.

[icybreeze]

Im good at maths *Mumbles i lieing * Heh sorry no can do pal" ^_^

[Eldin]

That's alright. This is a very hard proof.

[God of Twilight]

Originally Posted by Eldin

Any math whizzes out there? Could you help me prove that:

Cos (pi/5) = (1 + square root of 5)/4

Note that the second expression is the quantity of one plus the square root of five divided by four (hence the parenthesis).

Much appreciated.

You found someone!
Okay, gimme a minute...

?
I'm writing it out, and it looks wierd.
Are you allowed to graph or anything like that?

[icybreeze]

GOT-told you he's clever

Umm yeah i only posted to get first post,YAY.Im good at jokes people

[Eldin]

It's hard writing out the problem with the symbols on a keyboard.

Trust me. The equation is valid. I checked on my calculator. The goal is to get one side to equal the other by using various identities. I don't even know where to start, though.

[God of Twilight]

Originally Posted by Eldin

It's hard writing out the problem with the symbols on a keyboard.

Trust me. The equation is valid. I checked on my calculator. The goal is to get one side to equal the other by using various identities. I don't even know where to start, though.

Various identities?
I didnt really see anything that could use them.

The only thing I could do was change cos(pi/5) to cos36 degrees...
I'm not really sure what to do with the other side.
(I'm not even sure if changing from Radians to Degrees will do anything)

[Zien]

Cos (pi/5) = (1 + square root of 5)/4

CosXpie and CosX5= (I do not know) and (I do not know).
(I do not know)/(I do not know)=.25+2.23606798/4
(I do not know)/(I do not know)=.25+.559016995

I might of done it wrong, I don't know. Thats as far as I can go.

[God of Twilight]

Originally Posted by Zelda Ninja

Cos (pi/5) = (1 + square root of 5)/4

CosXpie and CosX5= (I do not know) and (I do not know).
(I do not know)/(I do not know)=.25+2.23606798/4
(I do not know)/(I do not know)=.25+.559016995

I might of done it wrong, I don't know. Thats as far as I can go.

That's not right...
You cant separate the (Pi/5)

[Zien]

Then would it be

(I do not know/I do not know) =.25+.559016995?

[God of Twilight]

Actually both sides are supposed to equal .809016994

Eldin, could you give examples of other problems that need to be solved in the same way, and show how you solved them?

[God of Twilight]

Okay Eldin, I'm not sure if you can do this, but:


What if you change cos(pi/5) to cos36 degrees.

Then you work the other side in order to get it's exact value, which is ~.80901

Next, you do the cos inverse of .80901, and you'll get cos36 degrees

Therefore, cos36=cos36.

- Will that work?

[Eldin]

Originally Posted by God of Twilight

Actually both sides are supposed to equal .809016994

This is true. That's the value I got.

Eldin, could you give examples of other problems that need to be solved in the same way, and show how you solved them?

This is an extra credit problem. It is far more advanced then any of the other problems I've done.

Basically, we've used the pythagorean identity (sin 2* + cos 2 = 1) and its variants, reciprocal identities, angle addition identities and so on. The class this is for is Trigonometry.

*Where 2 actually means squared.

What if you change cos(pi/5) to cos36 degrees.

It doesn't matter if it is in radians or degrees.

Will that work?

It technically works, but I don't think that's exactly what he's looking for.

[squall24]

What identities have you learned so far, I only know up to half-angle, so I might be able to help.

GOT: when you do identities you have to work on only one side of the equation.

[GDwarf]

Originally Posted by God of Twilight

Okay Eldin, I'm not sure if you can do this, but:


What if you change cos(pi/5) to cos36 degrees.

Then you work the other side in order to get it's exact value, which is ~.80901

Next, you do the cos inverse of .80901, and you'll get cos36 degrees

Therefore, cos36=cos36.

- Will that work?

No.

Proofs won't work if you convert anything to a non-finite decimal number or if you try to convert an irrational number into a finite decimal.

[God of Twilight]

Originally Posted by squall24

What identities have you learned so far, I only know up to half-angle, so I might be able to help.

GOT: when you do identities you have to work on only one side of the equation.

Actually you can do both.
(I would know, since I just did a chapter on identities last week)

[squall24]

Originally Posted by God of Twilight

Actually you can do both.
(I would know, since I just did a chapter on identities last week)

Maybe so, but it is best to work on one side only. Many teachers and professors want one side only proofs. It shows that you know how to do the steps fully. Trust me on this. My math professor told us about 75% of the math profs. on campus use the one side only approach.

[God of Twilight]

Originally Posted by squall24

Maybe so, but it is best to work on one side only. Many teachers and professors want one side only proofs. It shows that you know how to do the steps fully. Trust me on this. My math professor told us about 75% of the math profs. on campus use the one side only approach.

I realize that, but all I said was you CAN do both sides.
Anyways, this isnt the point of the thread...

[Project 2501]

EDIT: phail

[Eldin]

^Thanks. I figured we'd have to get those involved somewhere along the line. I'll take a closer look at this tomorrow to make sure it's right. If it is, I definitely owe you one.

[Project 2501]

Actually, I take it back; I based that on the idea that if your original answer was wrong, putting it through those formulas would produce a different value, which upon double-checking turned out not to be true; the "proof" is merely a tautology. Clearly I fail at logic today, so I went back and tried it again.

You can use angle complementation and half-angle formulas to make some progress;

cos[π/5] = +-√(1 + cos[2π/5]/2)

cos[2π/5] = sin[π/2 - 2π/5] = sin[π/10] = +-√(1 - cos[π/5]/2)

2(cos[π/5])^2 - 1 = cos[2π/5] = +-√(1 - cos[π/5]/2)

2(cos[π/5])^2 - 1 = +-√(1 - cos[π/5]/2)

8(cos[π/5])^4 - 8(cos[π/5])^2 + = -1 - cos[π/5]

which at least gets the entire thing in terms of cos[π/5].

Double edit: I can change it from a quartic into a cubic, but I doubt that the general solution for either is part of your coursework. The answer implies that it's possible to get the solution in quadratic form, but I can't see a way to do it. There is a way to get a sort-of proof, but I consider it semi-dishonest since the proof relies on plugging in the answer at a certain point (though I checked other answers this time and they don't check out; it will only work if the angle is correct, or with the correct angle times -1 as cosine an even function). I'll post it and let you see if you can divine anything from it, or pick out mistakes:

8(cos[π/5])^2[(cos [π/5])^2 - 1] = -1 - cos[π/5]

8(cos[π/5])^2 = -1 - cos[π/5]/{(cos[π/5])^2 - 1}

-8(cos[π/5])^2 = 1 + cos[π/5]/{cos[π/5] + 1}{cos[π/5] - 1}

8(cos[π/5])^2 = 1/(1 - cos[π/5])

8(sec[π/5])^2 = 1 - cos[π/5]

8(sec[π/5])^2 - 1 + cos[π/5] = 0

8/(cos[π/5])^2 - (cos[π/5])^2/(cos[π/5])^2 + (cos[π/5])^3/(cos[π/5])^2 = 0

8 + (cos[π/5])^3 - (cos[π/5])^2 = 0

(cos[π/5])^2 - (cos[π/5])^3 = 8

(cos[π/5])^2{1 - cos[π/5]} = 8

cos[π/5] = √(8/{1 - cos[π/5]})

At this point, you can substitute (1 + √5)/4 into the equation and it will check out. I've attempted to get a quadratic out of this jumble, but have met with no luck. All my attempts to set two equations for cos[π/5] equal to each other have resulted in aforementioned cubics.

[Soren 177]

I'm not sure if you still need help, but couldn't you use a triangle that has (pi/5) in it? Kind of like the right-angle triangle with the other 2 angles being (pi/4) and ((square root of 2)/2) for the adjacent and opposite and 1 as the hypotenuse.