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

[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?