Question- Suppose ABC is a triangle such that the lengths of AB, BC, CA are all rational numbers. Prove that, for all positive integer n, cos(nA) is a rational number.
My Reasoning: since cos(A) is rational because sides are rational so cos(nA) being a polynomial of cosine on expansion, it is too rational.
Is this a good proof, i would like you all to give some alternate better proofs for this... Thanks in advance
This is an immediate consequence of the law of cosines and of the multiplication formula. In fact $$\cos \gamma = \frac{a^2+b^2-c^2}{2ab},$$ and this proves the claim for $n=1$. The general statement is obtained by expanding $\cos (n \gamma)$ as a polynomial in $\cos \gamma$.