Why Focus Only on the Positive Root in Cosine Triple Angle Formula?

[Zetison]

Homework Statement

Find an albebraic expression for cos((1/3)*arccos(x)), in order to get rid of the trigonometric operands.

Homework Equations

cos(arccos(x))=x

The Attempt at a Solution

cos(x)= 4cos(x/3)^3 - 3cos(x/3), I can reduce cos(cos(x)/3)
by noting that cos(arccos(x))=x. Then I solve the cubic:
1 = 4y^3 - 3y
This was a tips I have found, but, it gave me nothing :P
Need help!

 

[Dick]

The equation you want to solve is x=4y^3-3y, isn't it? You want to solve for y in terms of x. Don't ask me about the algebra of solving the cubic. I really don't like solving cubics.

 

[Zetison]

Do you really need too use cubic to solve cos((1/3)*arccos(x))?
However, is it posible to get an albebraic expression for cos((1/3)*arccos(x)) without the trigonometric operands?

I can solve the cubic if you get it on the form ax^3+bx^2+cx+d=0 for me ;)

 

[g_edgar]

The equation you want to solve is x=4y^3-3y, isn't it? You want to solve for y in terms of x. Don't ask me about the algebra of solving the cubic. I really don't like solving cubics.

This is the answer to the problem. It is exactly solving a cubic.

 

[Hurkyl]

I don't know your teacher -- but "an algebraic expression for foo" does not necessarily mean that "foo" has to be isolated. It might be enough to simply have a purely algebraic equation involving foo.

 

[Dick]

Do you really need too use cubic to solve cos((1/3)*arccos(x))?
However, is it posible to get an albebraic expression for cos((1/3)*arccos(x)) without the trigonometric operands?

I can solve the cubic if you get it on the form ax^3+bx^2+cx+d=0 for me ;)

If you can solve cubics, then the problem is solve a*y^3+b*y^2+c*y+d=0, where a=4, b=0, c=-3 and d=(-x). (-x) is just a constant. I think Zetison may actually be expected to solve the cubic. There is already one post detailing an explicit cubic solution. I gave up half way through. I couldn't handle it. I did an explicit cubic solution once in my life. Because I REALLY wanted an expression for the real root. I hope never to have to do it again.

 

[LCKurtz]

If you can solve cubics, then the problem is solve a*y^3+b*y^2+c*y+d=0, where a=4, b=0, c=-3 and d=(-x). (-x) is just a constant. I think Zetison may actually be expected to solve the cubic. There is already one post detailing an explicit cubic solution. I gave up half way through. I couldn't handle it. I did an explicit cubic solution once in my life. Because I REALLY wanted an expression for the real root. I hope never to have to do it again.

Just for amusement I plugged that into Maple 10 and got this answer:

y = \frac{(x + \sqrt{-1+x^2})^{1/3}}{2} + \frac 1 {2(x+\sqrt{-1+x^2})^{1/3}}

 

[Zetison]

Just for amusement I plugged that into Maple 10 and got this answer:

y = \frac{(x + \sqrt{-1+x^2})^{1/3}}{2} + \frac 1 {2(x+\sqrt{-1+x^2})^{1/3}}

Makes no sense to me, because x is defined only in [-1,1]. While your equation is defined in (-infinite,-1]U[1,infinite)

 

[Ben Niehoff]

No, the expression is actually defined for all x. You get imaginary numbers, but the first term is the conjugate of the second term, so the imaginary parts cancel. When cos and arccos are analytically continued to the complex plane, the expression

\cos \arccos x = x

is true for all complex x (and in particular, for all real x).

 

[Zetison]

So with other words:

http://folk.ntnu.no/jonvegar/3
for all x and none irrational numbers?

 

[Zetison]

Maple says:
[PLAIN]http://folk.ntnu.no/jonvegar/images/capture.png

It did not work. I still need help!

 

[Cyosis]

That's because you're equating the wrong things. The correct solution is given in post #7 by LCKurtz.

What you've done in post #11 is the following

<br /> cos(1/3 \arccos(x))= 4\cos(x/3)^3 - 3\cos(x/3)<br />

which is wrong. Just read your own post again (post #1).

 

[Count Iblis]

No solution (without cheating like by using Maple) after 8 months?

 

[Count Iblis]

You can solve this problem easily as follows.

Using the identity:

cos(x) = [exp(i x) + exp(-i x)]/2

express arccos(x) in terms of logarithms.

Then compute cos[1/3 arccos(x)] using the obtained expression for arccos(x) and the above identity for cos(x).

 

[Zetison]

Hmmm, it shall be possible to write the equation without any logarithms.
If it is so easy, can you just write down the formula for me? I need it actually just to solv an other problem.

 

[Cyosis]

You've been given two methods. Either you solve the cubic equation or you use Count Iblis' method which is a lot faster. Perhaps it's time you show some work?

 

[Zetison]

Belive me, I have tried. But I don't understand the answer i get if I use the 2. method:

[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary.jpg

I want a answer without any imaginary parts. My x is defined only for [0,9], but when I cancel the imaginary part I do not get the right answer as demonstrated above...

 

[Cyosis]

Maybe you want to explain what you're doing instead of spitting out some maple math with some random numbers plugged in. I suggest once again that you use Count Iblis' method to derive the result instead of using some computer program.

 

[Zetison]

Can you define Count Iblis' method for me?

 

[Count Iblis]

Can you define Count Iblis' method for me?

You obtain an expression for y = arccos(x) by solving the equation:

[exp(i y) + exp(-iy)]/2 = x

Putting z = exp(iy), yields:

z + 1/z = 2 x

Can you take it from here?

 

[Zetison]

Ok, here is what I have done:
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary3.png
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary2.png
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary4.png
But when I'm testing my result in maple I get "false":
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary5.png

But even if it where true, how do I get an answer without imaginary parts?

 

[Zetison]

Is it really no one that can give me an answer to this problem??

 

[Cyosis]

You have been given the answer and multiple methods to obtain the answer. What more do you want?

 

[Zetison]

I want an answer to this problem:

Ok, here is what I have done:
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary3.png
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary2.png
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary4.png
But when I'm testing my result in maple I get "false":
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary5.png

But even if it where true, how do I get an answer without imaginary parts?

 

[guerom00]

Your answer is correct. Maple cannot handle the test you make

 

[Cyosis]

You're using a very inconvenient form of the arccos, which complicates everything. Just solve 2y=e^{i x}+e^{-i x} for x. Then you will get a pretty nice expression for the arccos and all i-s will vanish.

 

[Zetison]

How do you get an expression for x, when you got both e^ix and e^-ix? And even if I get an expression for arccos without i-s, I still have to get rid of cos...

 

[guerom00]

As far as I'm concerned the expresion you have fully answer your problem. All the rest is esthetics…

 

[Zetison]

No. The problem is not solved because i still have not got rid of the i-s. And that is not a trivial thing to do in my expression...

 

[guerom00]

No you don't. You stated your problem as getting rid of trigonometric functions. Mission accomplished Your expression with the i-s is more general since it extends to x<-1.

 

[Cyosis]

Zetison, this thread has been running for many months. Perhaps you should give a shot at what was suggested in post #26. Just make an appropriate substitution, z=e^{i x} and solve for z.

 

[Zetison]

Well, I can make a new thread where you have to get rid of the i-s...

This is not any schoolwork so I have all the time in the world

I have tried what's in post #26 with no result

How do you get an expression for x, when you got both eix and e-ix? And even if I get an expression for arccos without i-s, I still have to get rid of cos...

 

[Cyosis]

You will see it once you get there. Show your work and I will give you a hint to get past the point where you are stuck. Use the hint in post #31.

 

[Zetison]

2y=e^ix+e^-ix
z=e^ix
2y=z+z^-1
2yz=z²+1
z²-2yz+1=0
z=(y+(y²-1)^0,5) or z=(y-(y²-1)^0,5)
e^ix=(y+(y²-1)^0,5) or e^ix=(y-(y²-1)^0,5)

ix=ln(y+(y²-1)^0,5) or ix=ln(y-(y²-1)^0,5)

x=i^-1ln(y+(y²-1)^0,5) or x=i^-1ln(y-(y²-1)^0,5)

 

[Cyosis]

See you can do it! Take the solution with the positive root and plug it into the exponential form of your original equation.

 

[Zetison]

Yes, and this is what I get. I guess that's the far as I am going to get:
cos(\frac{1}{3}arccos(x)) = \frac{(x + \sqrt{x^2-1})^{1/3}}{2} + \frac 1 {2(x+\sqrt{x^2-1})^{1/3}}

But again, my x is defined as

x = \frac{473419}{6121\sqrt{6121}}

approximate x = 0.9885806704.

So my final goal here is to get exact values for

z = \frac{79}{60} + \frac{1}{30} \sqrt{6121} cos(\frac{1}{3}arccos(x)) = \frac{79}{60} + \frac{1}{30} \sqrt{6121} \frac{(x + \sqrt{x^2-1})^{1/3}}{2} + \frac 1 {2(x+\sqrt{x^2-1})^{1/3}}

But that is very difficult to express when
x = \frac{473419}{6121\sqrt{6121}}.

Is it possible?

 

[Zetison]

The problem of this thread is actually solved. Maybe I shall make a new thread about my main goal...

 

[Zetison]

See you can do it! Take the solution with the positive root and plug it into the exponential form of your original equation.

Why are we not interested in the negative root?