OPs integral can be written as $$ \int_0^{2\pi} f(50t) f(51t) \cos(2022 t) dt $$ with the auxiliary function

f[x_] := Sin[201*x]/Sin[x];

The division can be carried out explicitly, because (X^201-Y^201)/(X-Y) is a polynomial. The answer is $$ f(x) = e^{-200 ix} + e^{-198 ix} + \ldots + e^{-2ix} + 1 + e^{2ix} + \ldots + e^{198 ix} + e^{200 ix} $$ or as Mathematica code:

falternative[x_]:=Sum[Exp[I*k*x],{k,-200,200,2}];

To check this, use

f[x]-falternative[x]//FullSimplify
(* 0 *)

OPs integrand is therefore

integrand = falternative[50*t]*falternative[51*t]*Cos[2022*t]//TrigToExp//Expand;

This returns a linear combination of terms of the form $e^{i \omega t}$ with integer $\omega$, and the coefficient of $\omega = 0$ is $3$, therefore the value of the integral is $$ \int_0^{2\pi} f(50t) f(51t) \cos(2022 t) dt = 6\pi $$

Note: As OP has pointed out in a comment, a similar calculation is available here on Math SE.

OPs integral can be written as $$ \int_0^{2\pi} f(50t) f(51t) \cos(2022 t) dt $$ with the auxiliary function

f[x_] := Sin[201*x]/Sin[x];

The division can be carried out explicitly, because (X^201-Y^201)/(X-Y) is a polynomial. The answer is $$ f(x) = e^{-200 ix} + e^{-198 ix} + \ldots + e^{-2ix} + 1 + e^{2ix} + \ldots + e^{198 ix} + e^{200 ix} $$ or as Mathematica code:

falternative[x_]:=Sum[Exp[I*k*x],{k,-200,200,2}];

To check this, use

f[x]-falternative[x]//FullSimplify
(* 0 *)

OPs integrand is therefore

integrand = falternative[50*t]*falternative[51*t]*Cos[2022*t]//TrigToExp//Expand;

This returns a linear combination of terms of the form $e^{i \omega t}$ with integer $\omega$, and the coefficient of $\omega = 0$ is $3$, therefore the value of the integral is $$ \int_0^{2\pi} f(50t) f(51t) \cos(2022 t) dt = 6\pi $$