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I'm searching for a solution to the following functional equation:

$$f(u)f(u+lambda)=prod_i=1^Lrho(u-u_i)rho(u_i-u)+prod_i=1^Lrho(u+lambda-u_i)rho(u_i-u-lambda)$$

where $f$ is the unknown function, $rho(u)=sin(u-lambda)/sin(lambda)$, $lambda=pi/4$ and there are $L$ arbitrary real parameters $u_i_i=1^L$ where $L$ is an even number.

This equation comes from a $TQ$-equation for a solvable lattice model. If necessary, I can give more information but the equation itself is independent from its physical origin. I hope someone can help.

This is not a complete solution, but just presenting a few ideas.

Let's work on $rho$ first:
beginalign*
rho(u)&=fracsin(u-lambda)sin(lambda)\
&=fracsin(u)cos(lambda)-cos(u)sin(lambda)sin(lambda)\
&=sin(u)-cos(u),
endalign*
since $lambda=pi/4$ and $sin(pi/4)=cos(pi/4).$

Next, we look at what happens when we have $rho(u)cdotrho(-u).$ We have
beginalign*
rho(u)cdotrho(-u)&=[sin(u)-cos(u)][-sin(u)-cos(u)] \
&=-sin^2(u)-sin(u)cos(u)+sin(u)cos(u)+cos^2(u)\
&=cos^2(u)-sin^2(u)\
&=cos(2u).
endalign*

From this, we gather that
beginalign*f(u),f(u+lambda)&=prod_i=1^Lrho(u-u_i),rho(u_i-u)+prod_i=1^Lrho(u+lambda-u_i),rho(u_i-u-lambda)\
&=prod_i=1^Lcos(2(u_i-u))+prod_i=1^Lcos(2(u_i-u-lambda)) \
&=prod_i=1^Lcos(2(u_i-u))+prod_i=1^Lcos(2(u_i-u)-pi/2),;textor \
f(u),f(u+pi/4)&=prod_i=1^Lcos(2(u_i-u))+prod_i=1^Lsin(2(u_i-u)).
endalign*
From this, we can at least see that if $u=u_i$ for any $1le ile L,$ then the second product drops out and you get
$$f(u_i),f(u_i+pi/4)=prod_beginarraycj=1\ jnot=iendarray^Lcos(2(u_j-u)).$$
Conversely, if $u_i-u=(2k+1)pi/4$ for some $kinmathbbZ,$ then the first product series drops out and you only get the second one.

Finally, it's always worthwhile plugging in $u=0$ to arrive at
$$f(0),f(pi/4)=prod_i=1^Lcos(2u_i)+prod_i=1^Lsin(2u_i).$$

The RHS of
$$f(u),f(u+pi/4)=prod_i=1^Lcos(2(u_i-u))+prod_i=1^Lsin(2(u_i-u))$$
is $pi$ periodic, and hence the LHS is as well.