# How do you simplify cos((1/3)arccos((1/3)x))?

Oct 29, 2016

This is already in simplest form, unless you would consider the following simpler:

$\cos \left(\left(\frac{1}{3}\right) \arccos \left(\frac{x}{3}\right)\right)$

$= \frac{1}{2} \left({\left(\frac{x}{3} + i \sqrt{1 - {\left(\frac{x}{3}\right)}^{2}}\right)}^{\frac{1}{3}} + {\left(\frac{x}{3} - i \sqrt{1 - {\left(\frac{x}{3}\right)}^{2}}\right)}^{\frac{1}{3}}\right)$

#### Explanation:

Expressions of the form $\cos \left(\frac{1}{3} \arccos \left(\frac{1}{3} x\right)\right)$ are encountered when solving cubic equations using trigonometric substitution.

For example, consider $x = 3 \cos \left(\frac{\pi}{3}\right) = \frac{3}{2}$.

Then:

$\cos \left(\left(\frac{1}{3}\right) \arccos \left(\left(\frac{1}{3}\right) x\right)\right) = \cos \left(\frac{1}{3} \arccos \left(\frac{1}{2}\right)\right) = \cos \left(\frac{\pi}{9}\right)$

which has no expression in terms of Real $n$th roots.

We can express something in terms of a Complex cube root:

Using de Moivre's Theorem:

$\cos \left(\left(\frac{1}{3}\right) \arccos \left(\frac{x}{3}\right)\right) + i \sin \left(\left(\frac{1}{3}\right) \arccos \left(\frac{x}{3}\right)\right)$

$= {\left(\cos \arccos \left(\frac{x}{3}\right) + i \sin \arccos \left(\frac{x}{3}\right)\right)}^{\frac{1}{3}}$

$= {\left(\frac{x}{3} + i \sqrt{1 - {\left(\frac{x}{3}\right)}^{2}}\right)}^{\frac{1}{3}}$

Similarly:

$\cos \left(\left(\frac{1}{3}\right) \arccos \left(\frac{x}{3}\right)\right) - i \sin \left(\left(\frac{1}{3}\right) \arccos \left(\frac{x}{3}\right)\right)$

$= {\left(\cos \arccos \left(\frac{x}{3}\right) - i \sin \arccos \left(\frac{x}{3}\right)\right)}^{\frac{1}{3}}$

$= {\left(\frac{x}{3} - i \sqrt{1 - {\left(\frac{x}{3}\right)}^{2}}\right)}^{\frac{1}{3}}$

We can add these two results and halve the sum to find:

$\cos \left(\left(\frac{1}{3}\right) \arccos \left(\frac{x}{3}\right)\right)$

$= \frac{1}{2} \left({\left(\frac{x}{3} + i \sqrt{1 - {\left(\frac{x}{3}\right)}^{2}}\right)}^{\frac{1}{3}} + {\left(\frac{x}{3} - i \sqrt{1 - {\left(\frac{x}{3}\right)}^{2}}\right)}^{\frac{1}{3}}\right)$