# Solve cos((1/3)*arccos(x))

1. Sep 10, 2009

### Zetison

1. The problem statement, all variables and given/known data
Find an albebraic expression for cos((1/3)*arccos(x)), in order to get rid of the trigonometric operands.

2. Relevant equations
cos(arccos(x))=x

3. 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!

2. Sep 10, 2009

### 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.

3. Sep 10, 2009

### 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 ;)

4. Sep 10, 2009

### g_edgar

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

5. Sep 10, 2009

### Hurkyl

Staff Emeritus
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.

6. Sep 10, 2009

### Dick

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.

Last edited: Sep 11, 2009
7. Sep 11, 2009

### LCKurtz

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

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

8. Sep 11, 2009

### Zetison

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

Last edited: Sep 11, 2009
9. Sep 11, 2009

### 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).

10. Sep 11, 2009

### Zetison

So with other words:

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

Last edited by a moderator: May 4, 2017
11. May 18, 2010

### Zetison

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

It did not work. I still need help!

Last edited by a moderator: May 4, 2017
12. May 18, 2010

### 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

$$cos(1/3 \arccos(x))= 4\cos(x/3)^3 - 3\cos(x/3)$$

13. May 18, 2010

### Count Iblis

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

14. May 18, 2010

### 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).

15. May 18, 2010

### 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.

16. May 19, 2010

### 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?

17. May 19, 2010

### 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 [Broken]

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...

Last edited by a moderator: May 4, 2017
18. May 19, 2010

### 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.

Last edited: May 19, 2010
19. May 19, 2010

### Zetison

Can you define Count Iblis' method for me?

20. May 19, 2010

### Count Iblis

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?