Find $\cos 8x$ if:
$$\tan x \tan 2x = \cot 2x \cot 3x$$
We can verify quickly that $\tan 2x \to \infty$ and $\tan 3x \to \infty$ are not solutions of the trig equation, so the equation may be rewritten as:
$$\tan x \tan^2 2x \tan 3x = 1$$
Using expansion formulas, we may see that:
$$\tan 2x = \frac{2 \tan x}{1 – \tan^2 x}$$
And also:
$$\tan 3x = \frac{3 \tan x – \tan^3 x}{1 – 3 \tan^2 x}$$
By long calculations, I have obtained that:
$$\tan x \in \{-1 – \sqrt{2}, -1 + \sqrt{2}, 1 – \sqrt{2}, 1 + \sqrt{2}\}$$
However, I am not able to take those calculations further. I have learnt about Chebyshev polynomials, but using WolframAlpha, the expression is really hairy and not realy easy to cope with.
Is there a smarter trick to solve the question? If not, how should I simplify my calculations?
Best Answer
Let $t:=\tan x$ and $T:=t^2\ge0$ (assuming $x\in\Bbb R$), so$$\frac{4T^2(3-T)}{(1-T)^2(1-3T)}=1\implies (T+1)(T^2-6T+1)=0\implies T^2-6T+1=0,$$and$$\cos4x=2\left(\frac{1-T}{1+T}\right)^2-1=\frac{T^2-6T+1}{(1+T)^2}=0\implies\cos8x=-1.$$