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Alright, so I am told to prove that:

$$\tan (3A) = \frac{3\tan(A)-\tan^3(A)}{1-3\tan^2(A)}$$

This can be pretty easily done by applying the $\tan$ addition formula, taking the angles $2A$ and $A$, upon which one then applies the $\tan$ double angle formula. Simplifying the resultant mess indeed yields the above identity. To show that I have attempted the question:

$$\tan(2A + A) = \frac{\tan(2A) + \tan(A)}{1-\tan(2A)\tan(A)}$$

Note:

$$\tan(2A)= \frac{2\tan(A)}{1-\tan^2(A)}$$

Substituting:

$$\tan(3A)=\frac{2\tan(A)/(1-\tan^2(A))+\tan(A)}{1-(2\tan(A)/(1-\tan^2(A)))\tan(A)}$$

Which simplifies to:

$$\tan(3A)=\frac{(3tan(A)-tan^3(A))/(1-tan^2(A))}{(1-tan^2(A))/(1-3tan^2(A))}$$

The two $(1-\tan^2(A))$ cancel out and we attain our desired identity..

However, I am told to make use of the fact that the cubic polynomial

$$t^3 - 3t^2 - 3t +1 $$

factorizes into:

$$(t+1)(t^2-4t+1)$$

Clearly, the polynomial and the trig identity look similar: i.e. $3\tan = 3t$ or $1-3t^2 = 1 - 3\tan^2$, etc.

However, how can the factorization aid the proof? I am currently at a brain block, so a clue would be very much appreciated! :)

Additionally, this should not really require any use of complex numbers / De Moivre as this question comes from a chapter in which complex numbers have not been introduced.

Thank you!

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You have a typo in your very first line. $\tan (3A) = \frac{3\tan(A)-\tan^{\color{red}{2}}(A)}{1-3\tan^2(A)}$ should be changed to $\tan (3A) = \frac{3\tan(A)-\tan^{\color{blue}{3}}(A)}{1-3\tan^2(A)}$. –  David H May 31 at 20:50

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