Solve $\lambda=\dfrac{\sin(w \mathrm{t} +\frac{w}{2})}{\sin(\frac{w}{2})}$ Is it possible to write $w$ as a function of $\lambda$ for the following equation ?
$$
\lambda=\dfrac{\sin(w \mathrm{t} +\frac{w}{2})}{\sin(\frac{w}{2})}
$$
Where  $t $ is an integer >1.
if there was only one term with sinus, it's easy to find, but I do not see how to proceed with this one.
Any help is welcomed.
 A: It is useful to recall that the Chebyshev polynomials of the 2nd kind satisfy the functional equation $$U_n(\cos \theta)=\frac{\sin((n+1)\theta)}{\sin\theta}.$$
That is, the ratio on the RHS can be written as a degree-$n$ polynomial in $\cos\theta$. Consequently we may express the equation of interest as $$\lambda =\frac{\sin((w/2)(2t+1))}{\sin(w/2)} = U_{2t}(\cos(w/2)).$$ (This bears out Michael Seifert's comment above.) Hence to solve for $\cos(w/2)$ we'd want to express the roots of a degree-$(2t)$ polynomial as a function of $\lambda$. But this can't be done in closed-form in any useful sense (e.g. no general solution for a quintic polynomial), so one has to be satisfied with numerical methods.
A: No, there is no closed-form solution for $w$ in general.  
EDIT: If $t$ is a positive integer, $$\frac{\sin((t+1/2) w)}{\sin(w/2)} = U_{2t}(\cos(w/2))$$
where $U_{n}$ are the Chebyshev polynomials of the second kind.
$U_{2t}$ is a polynomial of degree $2t$ whose roots are all in the interval $[-1,1]$, the greatest root being $\cos(\pi/(2t+1))$, and $U_{2t}(1) = 2t+1$.  On the interval $[a, 1]$ where $a = \cos(\pi/(2t+1))$, we might approximate
$U_{2t}(x)$ by a quadratic
$$ U_{2t}(x) \approx \frac{(x-a)(2t+1)}{1-a} + \frac{2 (x-a)(1-x) (2 m - 2t - 1)}{(1-a)^2} $$
where $m = U_{2t}((1+a)/2)$.  Solve that quadratic to get an approximate solution.  
Of course you could just use Newton's method...
A: i think this is only possible with a numerical method i have
$$\frac{\sin(wt)\cos(w/2)+\cos(wt)\sin(w/2)}{\sin(w/2)}=\sin(wt)\cot(w/2)+\cos(wt)=\Lambda$$
