(Beginner) Intuition to solve a functional equation and steps for this particular- Please can someone please tell me the intuition behind solving a functional equation? 
For example, $$f(x)+f(y)=2f\left(\frac{x+y}2\right)f'\left(\frac{x-y}2\right)$$
Now at the first look, it seems like $f(x)=\sin x$, but how to solve it rigorously?

P.S: As written in the title, I am a beginner and I have no experience at solving a functional equation. I know little bit about series expansions and I know calculus upto 12th grade. (Derivatives, Integrals (Impropers-a bit), Limits)

$$f(x)+f(y)=2f\left(\frac{x+y}2\right)f'\left(\frac{x-y}2\right)$$
Let's define
$t=\frac{x+y}{2}$,
$u=\frac{x-y}{2}$
$$f(t+u)+f(t-u)=2f\left(t\right)f'\left(u\right)$$
for
$u=0$
$$f(t)+f(t)=2f\left(t\right)f'\left(0\right)$$
$$2f(t)=2f\left(t\right)f'\left(0\right)$$
$$f'\left(0\right)=1$$
To derivate both side for $u$
$$f'(t+u)-f'(t-u)=2f\left(t\right)f''\left(u\right)$$
Then 
for
$u=0$ again
$$f'(t)-f'(t)=2f\left(t\right)f''\left(0\right)$$
$$0=2f\left(t\right)f''\left(0\right)$$
$$f''\left(0\right)=0$$
To derivate both side for $u$ again
$$f''(t+u)+f''(t-u)=2f\left(t\right)f'''\left(u\right)$$
Then 
for
$u=0$ again
$$2f''(t)=2f\left(t\right)f'''\left(0\right)$$
$$f'''\left(0\right)=c^2$$
if $c=0$ then
$$f''(t)=0$$ and
$$f(t)=a+bt$$ and
 we know that
$$f'\left(0\right)=1$$
Thus $b=1$
$$f(t)=a+t$$ 
to put into the function equation to find $a$
$$a+x+a+y=2(a+\frac{x+y}2)$$
$$2a+x+y=2a+x+y$$
thus
$$f(t)=a+t$$  is general solution for  $c=0$ where $a∈C$ .
if $c \neq0$ then
$$f''(t)=c^2f\left(t\right)$$
The solution of the equation 
$$f(t)=k_1e^{ct}+k_2e^{-ct}$$
we know 
$$f'\left(0\right)=1$$
$$f''\left(0\right)=0$$ and 
$$f'''\left(0\right)=c^2$$
To solve all
$$f'(t)=k_1ce^{ct}-k_2ce^{-ct}$$
$$f'(0)=c(k_1-k_2)$$
$$c(k_1-k_2)=1$$

$$f''(t)=k_1c^2e^{ct}+k_2c^2e^{-ct}$$
$$f''(0)=c^2(k_1+k_2)$$
$$c^2(k_1+k_2)=0$$
$$k_1=-k_2$$
Thus
$$c(k_1-k_2)=1$$
$$2ck_1=1$$
$$k_1=\frac{1}{2c}$$
$c \neq 0$
Our general solution is for  $c \neq 0$
$$f(t)=\frac{1}{2c}(e^{ct}-e^{-ct})$$
To confirm the solution
$$\frac{1}{2c}(e^{cx}-e^{-cx})+\frac{1}{2c}(e^{cy}-e^{-cy})=2\frac{1}{2c}(e^{c\frac{x+y}2}-e^{-c\frac{x+y}2})\frac{1}{2}(e^{c\frac{x-y}2}+e^{-c\frac{x-y}2})$$
$$\frac{1}{2c}(e^{cx}-e^{-cx}+e^{cy}-e^{-cy})=\frac{1}{2c}(e^{c\frac{x+y}2}-e^{-c\frac{x+y}2})(e^{c\frac{x-y}2}+e^{-c\frac{x-y}2})$$
$$\frac{1}{2c}(e^{cx}-e^{-cx}+e^{cy}-e^{-cy})=\frac{1}{2c}(e^{cx}+e^{cy}-e^{-cy}-e^{-cx})$$
and you are right that $c=i$ ,$f(t)=\sin(t)$ is a solution of the function equation but not general solution. it is just particular solution
$$f_i(t)=\frac{1}{2i}(e^{it}-e^{-it})=\frac{1}{2i}(\cos{}t +i\sin{t} -\cos{t}+i \sin{t})=\sin{t}$$
The general formula is :
for $c \neq 0$;
$$f(t)=\frac{1}{2c}(e^{ct}-e^{-ct})$$
for  $c = 0$;
$$f(t)=a+t$$  ;where $a∈C$   
and also $f(t)=0$ is a solution of the function equation.
