# Exact arctangent of product of tangents

Calculate $x$, if $$\tan(x)=\tan9\tan69\tan33$$

(Using sexagesimal degrees) Since $\tan3x=\tan(60-x)\tan x \tan(60+x)$: \begin{align*} \tan27&=\tan69\tan9\tan51\\ \implies\tan27\tan39&=\tan69\tan9 \end{align*}

So the problem is equivalent to calculating $x$ in $$\tan(x)=\tan27\tan33\tan39$$ But thats all my progress so far. Interestingly enough, the answer is $x=15$. Is there some way tu constructively solve the equation? If not, a straightforward proof of $\tan(15)=\tan9\tan69\tan33$ would be nice too.

• Hint: $27=60-33$. – Lucian Jan 8 '15 at 21:24
• another hint: $33 = 30 + 3$ – John Joy Jan 8 '15 at 21:36
• @Lucian So $\tan99=\tan27\tan33\tan93\implies \tan81\tan3=\tan27\tan33$. I don't see what to do next. – chubakueno Jan 8 '15 at 22:26
• upload.wikimedia.org/math/e/0/5/… This may help – Teoc Jan 11 '15 at 2:35
• With tangent addition formulae you can compute the closed forms of the tangents. – Teoc Jan 11 '15 at 2:36

$$\tan(x)=\tan(9)\tan(69)\tan(33)$$ $$\tan(x) = \frac{\tan(39)\tan(3)}{\tan(9)}$$ Thus, $$\tan^2(x) = \tan(3)\tan(33)\tan(39)\tan(69)$$ $$\tan^2(x) = \tan(3)\tan(3-36)\tan(3-72)\tan(3+36)$$ $$\tan^2(x)\tan(75) = \tan(3-72)\tan(3-36)\tan(3)\tan(3+36)\tan(3+72)$$
$$\tan(5x)= \tan(x-72)\tan(x-36)\tan(x)\tan(x+36)\tan(x+72)$$
Let $z=\cos(x)+i\sin(x)$, and $\omega = \cos(36)+i\sin(36)$ Then, $$\tan(x) = -\frac{i(z^2-1)}{2(z^2+1)}$$ Similarly we can get $\tan(x-36),\tan(x-72),\tan(x+36),\tan(x+72)$ and then multiply . We very easily get the above identity. So, we have $$\tan^2(x)=\tan^2(15) \implies x=15$$
• Where did you use $\omega$? – user84413 Jan 18 '15 at 2:10