Link between a cubic polynomial and a trig identity 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:
$$\tan(3A)=\frac{(3\tan(A)-\tan^3(A))/(1-\tan^2(A))}{(1-\tan^2(A))/(1-3\tan^2(A))}$$
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!
 A: It's a lot of work to proof the identity so ;) :
Verify the following identity:
$$\tan(3a)=\frac{3\tan(a)-\tan^3(a)}{1-3\tan^2(a)}=>$$
$$\tan(3a)\left(1-3\tan^2(a)\right)=3\tan(a)-\tan^3(a)=>$$
$$\left(1-3\left(\frac{\sin(a)}{\cos(a)}\right)^2\right)\frac{\sin(3a)}{\cos(3a)}=3\frac{\sin(3a)}{\cos(3a)}-\left(\frac{\sin(a)}{\cos(a)}\right)^3=>$$
$$\frac{\sin(3a)\left(1-\frac{3\sin^2(a)}{\cos^2(a)}\right)}{\cos(3a)}=3\left(\frac{\sin(a)}{\cos(a)}\right)-\left(\frac{\sin(a)}{\cos(a)}\right)^3=>$$
$$\frac{\sin(3a)\left(1-\frac{3\sin^2(a)}{\cos^2(a)}\right)}{\cos(3a)}=\frac{3\sin(a)}{\cos(a)}-\frac{\sin^3(a)}{\cos^3(a)}=>$$
$$\frac{\sin(3a)}{\cos(3a)}\frac{\cos^2(a)-3\sin^2(a)}{\cos^2(a)}=\frac{3\sin(a)}{\cos(a)}-\frac{\sin^3(a)}{\cos^3(a)}=>$$
$$\frac{\sin(3a)\left(\cos^2(a)-3\sin^2(a)\right)}{\cos^2(a)\cos(3a)}=\frac{3\cos^2(a)\sin(a)-\sin^3(a)}{\cos^3(a)}=>$$
$$\cos^3(a)\sin(3a)\left(\cos^2(a)-3\sin^2(a)\right)=\cos^2(a)\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)sin(3a)\left(\cos^2(a)-3\sin^2(a)\right)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)\sin(3a)\left(1-\sin^2(a)-3\sin^2(a)\right)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)\left(1-4\sin^2(a)\right)\sin(3a)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)\left(1-4\sin^2(a)\right)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)\left(1-4\sin^2(a)\right)\left(3(1-\sin^2(a))\sin(a)-\sin^3(a)\right)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$\cos(a)\left(3\sin(a)-4\sin^3(a)\right)\left(1-4\sin^2(a)\right)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=\cos(3a)\left(3\cos^2(a)\sin(a)-\sin^3(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=\cos(3a)\left(3(1-\sin^2(a))\sin(a)-\sin^3(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=\cos(3a)\left(3\sin(a)-3\sin^3(a)-\sin^3(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=\left(\cos^3(a)-3\cos(a)\sin^2(a)\right)\left(3\sin(a)-3\sin^3(a)-\sin^3(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=\left(3\sin(a)-4\sin^3(a)\right)\left(\cos(a)-4\cos(a)\sin^2(a)\right)=>$$
$$3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)=3\cos(a)\sin(a)-16\cos(a)\sin^3(a)+16\cos(a)\sin^5(a)$$
The left hand side and riht hand side are identical, identity has been verified
