<s>[INST] How to simplify $\cos(\frac{1}{3}\cos^{-1}(x))$ and $\sin(\frac{1}{3}\cos^{-1}(x))$? How to simplify this trigonometric expression
$$
\cos \left(\frac{1}{3} \cos ^{-1}(b)\right)
$$
I want to write this equation as
$$
\cos \left(\frac{1}{3} \cos ^{-1}(b)\right) = \cos(\cos^{-1}(B)) = B
$$
Where $B$ is written in terms of $b$.
I couldn't find any identity that would let me do it.
This term appears in the solution of the depressed cubic equation as given in wikipedia.
In the original equation, the term is like
$$
\cos\left(\frac{1}{3}\cos^{-1}(b) - \frac{2\pi n}{3}\right) \\
= \cos\left(\frac{1}{3}\cos^{-1}(b)\right)\cos\left(\frac{2\pi n}{3}\right)+\sin\left(\frac{1}{3}\cos^{-1}(b)\right)\sin\left(\frac{2\pi n}{3}\right)
$$
Here the only term that are bothering me are only those $\cos\left(\frac13\cos^{-1}(b)\right)$ and $\sin\left(\frac13\cos^{-1}(b)\right)$
 [/INST] Use trig. identity $$4\cos^3x-3\cos x=\cos 3x$$
substitute $x=\frac13\cos^{-1}b$
$$4\cos^3\left(\frac13\cos^{-1}b\right)-3\cos\left(\frac13\cos^{-1}b\right)=b$$
$$4B^3-3B-b=0$$
Solve above cubic equation for $B$ i.e. $\cos\left(\frac13\cos^{-1}b\right)$
</s>