TITLE: When is $\frac1{n}\sin(nt)=\frac1{n+2}\sin((n+2)t)$?
QUESTION [5 upvotes]: I'm trying to classify (at least as fully as possible) the values of $t$ in $(0,\pi/2)$ for which the following equation has a solution for some natural number, $n$.
$$\frac1{n}\sin(nt)=\frac1{n+2}\sin((n+2)t)$$
Does anyone know of any results that will help me?

REPLY [2 votes]: EDIT: See J.M.'s comment to Rahul's answer: it's better to use the "other" Chebyshev polynomials.
Square your identity: $$(n+2)^2 \sin^2 nt  = n^2 \sin^2 (n+2)t.$$ Convert from sines to cosines: $$(n+2)^2 + n^2 \cos^2 (n+2)t = n^2 + (n+2)^2 \cos^2 nt.$$ Now let $x = \cos t$ to get $$(n+2)^2 + n^2 T_{n+2}(x)^2 = n^2 + (n+2)^2 T_n(x)^2,$$ where the $T_m$ are the Chebyshev polynomials. We get a polynomial equation of degree $2(n+2)$ in $x$ which you can solve to find all solutions (you will need to check that the signs of the sines match).