Zetison
- 35
- 0
Homework Statement
Find exact values for z without any trigonometrical components or imaginary numbers
x = \frac{79}{60} + \frac{1}{30} \sqrt{6121} cos\left(\frac{1}{3}arccos(z)\right)
where
z = \frac{473419}{6121\sqrt{6121}}
Homework Equations
cos\left(\frac{1}{3}arccos(z)\right) = \frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}}
The Attempt at a Solution
I pretty fast get imaginary numbers from the equation above. But they shall, as I know, cancel one way or the other. So this is what I get:
x = \frac{79}{60} + \frac{1}{30} \sqrt{6121} cos\left(\frac{1}{3}arccos(z)\right) = \frac{79}{60} + \frac{1}{30} \sqrt{6121} \frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}}
x = <br /> \frac{79}{60} + \frac{1}{30} \sqrt{6121} \left(\frac{\left({ \frac{473419}{6121\sqrt{6121}} + \sqrt{ -\frac{5207760000}{229333309561}}}^{1/3}}{2} + <br /> \frac {1}{2\left({ \frac{473419}{6121\sqrt{6121}} + \sqrt{ -\frac{5207760000}{229333309561}}}\right)^{1/3}}\right)x = <br /> \frac{79}{60} + \frac{1}{60} \sqrt{6121} \left(\left({ \frac{473419}{6121\sqrt{6121}} + i \sqrt{\frac{5207760000}{229333309561}}}\right)^{1/3} + <br /> \frac 1 {\left({ \frac{473419}{6121\sqrt{6121}} + i \sqrt{\frac{5207760000}{229333309561}}}\right)^{1/3}}\right)x = <br /> \frac{79}{60} + \frac{1}{60} \left({473419 + 600 i \sqrt{14466}}\right)^{1/3} + <br /> \frac {6121}{60\left({473419 + 600 i \sqrt{14466}}\right)^{1/3}}
Let
u = 473419 + 600 i \sqrt{14466}
Then I have
x = <br /> \frac{79}{60} + \frac{1}{60} {u}^{1/3} + <br /> \frac {6121}{60{u}^{1/3}}
x = <br /> \frac{79}{60} + \frac{{u}^{2/3} + 6121}{<br /> {60{u}^{1/3}}}
x = <br /> \frac{79}{60} + \frac{{u}^{4/3} + 6121{u}^{2/3}}{<br /> {60u}}But from here, I can not see how I can get rid of the i-s...
Last edited: