I found the following question in a book without any proof:

Question : Suppose that each length of three edges of a triangle $ABC$ are $BC=a, CA=b, AB=c$ respectively. If $$\frac1a=\frac1b+\frac1c, \frac2a=\frac{1}{c-b}-\frac{1}{c+b},$$ then prove $$\sqrt[3]{\cos{2A}}+\sqrt[3]{\cos{2B}}+\sqrt[3]{\cos{2C}}=\sqrt[3]{\frac{5-3\color{blue}{\sqrt[3]{7}}}{2}}\ .$$

This book says that this is the question by Ramanujan.

I've tried to prove this, but I'm facing difficulty. Can anyone help? If you have any helpful information, please let me know it.

  • $\begingroup$ I've corrected the typo (in blue). $\endgroup$ – Tito Piezas III Feb 6 '15 at 9:04


$$x^3 + x^2 - (3 n^2 + n)x + n^3=0\tag1$$


$$F_p = x_1^{1/3}+x_2^{1/3}+x_3^{1/3} = \sqrt[3]{-(6n+1)+3\sqrt[3]{np}}\tag2$$

where $p = 9n^2+3n+1$. This is a special case of an identity by Ramanujan. Let $n=-1$, change variables $x \to t$, and we have Lucian's,



$$t_1^{1/3}+t_2^{1/3}+t_3^{1/3} = \sqrt[3]{5-3\sqrt[3]{7}}$$

We deduce,

$$t_1 = 2\cos 2A,\quad t_2 = 2\cos 2B,\quad t_3 = 2\cos 2C$$



correcting a typo in the original question. (It should be $\color{blue}{\sqrt[3]{7}}$.)

  • $\begingroup$ naive question, how do you know that the absolute value of $t_1, t_2, t_3$ are less than or equal to $2.$ $\endgroup$ – abel Dec 23 '14 at 4:49
  • $\begingroup$ @abel I'm afraid I won't be able to give a rigorous answer to that. I just recognized that this was a particular example of a special case of a general cubic identity by Ramanujan. $\endgroup$ – Tito Piezas III Dec 23 '14 at 4:55
  • $\begingroup$ you can show, simply by dividing by $t-2$ and $t+2$ that all three roots, one positive and two negative, is contained in $[-2,2$. $\endgroup$ – abel Dec 23 '14 at 5:07

$$\begin{align}\text{Partial answer:}\qquad\qquad\quad\frac1a=\ \frac1b+\frac1c\quad\iff\quad&a=\frac1{\frac1b+\frac1c}=\frac{bc}{b+c}\qquad\qquad(1)\\\\\\\\\\\frac2a=\frac1{c-b}-\frac1{c+b}\quad\iff\quad\frac2a=\frac{2b}{c^2-b^2}\quad\iff\quad&a=\frac{c^2-b^2}b\ \qquad\qquad\qquad\quad\ (2)\end{align}$$

$$\begin{align}&\overset{(1)}{\underset{(2)}\iff}\quad\frac{bc}{b+c}=\frac{c^2-b^2}b\quad&\iff&\quad b^2c=(c-b)(b+c)^2\ \qquad\ |:(b^2c)\\\\\\\\&\iff\quad1=\frac{c-b}c\left(\frac{b+c}b\right)^2\quad&\iff&\quad1=\left(1-\frac1t\right)(1+t)^2\qquad|\cdot t\end{align}$$

$$\iff\quad t=(t-1)(t+1)^2\quad\iff\quad t^3+t^2-2t-1=0\quad\iff\quad t=\frac cb=\ldots$$

We are dealing with a so-called irreducible case , with a single positive real root. Furthermore, from the Generalized Pythagorean Theorem, also known as the Law of Cosines, we deduce the following three relationships, and their implications:

$$a^2+b^2-2ab\cos C=c^2\iff\cos C=\frac{a^2+b^2-c^2}{2ab}\underset{(2)}=\frac{a^2-ab}{2ab}=\frac12\left(\frac ab-1\right)\underset{(2)}=\frac{t^2-1}2$$

$$a^2+c^2-2ac\cos B=b^2\iff\cos B=\frac{a^2+c^2-b^2}{2ac}\underset{(2)}=\frac{a^2+ab}{2ac}=\frac{a+b}{2c}\ \underset{(2)}=\ \frac c{2b}=\frac t2$$

$$b^2+c^2-2bc\cos A=a^2\iff\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac12\left(\frac bc+\frac cb-\frac{a^2}{bc}\right)\underset{(1)}=\underset{(1)}=\frac12\left(\frac bc+\frac cb-\frac{a}{b+c}\right)\underset{(2)}=\frac12\left(\frac bc+\frac cb-\frac{c-b}b\right)=\frac12\left(\frac bc+1\right)=\frac12\left(\frac1t+1\right)$$

$$\text{Then we use:}\qquad\cos2x=\cos^2x-\sin^2x=\cos^2x-(1-\cos^2x)=2\cos^2x-1$$

  • $\begingroup$ There a general formula to find Ramanujan-type identities with similar simplicity to the one above. Kindly see this updated answer. $\endgroup$ – Tito Piezas III Jan 4 '15 at 5:52

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