(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)
 A: $$f(x)+f(y)=2f\left(\frac{x+y}2\right)f'\left(\frac{x-y}2\right)$$
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.
A: Taking $x=y$ gives
$$ 2f(x) = 2f(x)f'(0), \tag{1} $$
so either $f(x) \equiv 0$ or $f'(0)=1$. Suppose it is the latter (else we have found $f$). Setting $x=-y$ gives
$$ f(x)+f(-x) = 2f(0)f'(x). \tag{2} $$
Differentiating the original equation with respect to $x$ gives
$$ f'(x) = f'\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right) + f\left( \frac{x+y}{2} \right)f''\left( \frac{x-y}{2} \right), \tag{3} $$
and setting $y=-x$ gives
$$ f'(x) = f'(0)f'(x)+f(0)f''(x) = f'(x)+f(0)f''(x). $$
This says that either $f''(x) \equiv 0$ or $f(0)=0$. It is easy to check that the only functions with $f''(x) \equiv 0$ which satisfy the equation are $f(x) =0$ and $f(x)=x$, so now assume $f(0)=0$. (2) then gives that
$$ f(-x) = -f(x), $$
so we also have $f''(0)=0$. Differentiating (3) with respect to $y$ gives
$$ 0 = f''\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right) - f'\left( \frac{x+y}{2} \right)f''\left( \frac{x-y}{2} \right) + f\left( \frac{x+y}{2} \right)f''\left( \frac{x-y}{2} \right) - f\left( \frac{x+y}{2} \right)f'''\left( \frac{x-y}{2} \right) \\
= f''\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right) - f\left( \frac{x+y}{2} \right)f'''\left( \frac{x-y}{2} \right). $$
Now, if we take $y=x$, this reduces to
$$ f'''(0) f(x) = f''(x) f'(0) = f''(x). $$
At this point, we can write down the general solution to this equation: setting $f'''(0)=A^2$ gives
$$ f(x) = B\sinh{Ax}+C\cosh{Ax}, $$
and imposing the boundary conditions $f(0)=0$, $f'(0)=1$ gives general solution
$$ f(x) = \frac{1}{A}\sinh{Ax}. $$
It is simple to check that this always satisfies the functional equation, and further, you can recover the $f(x)=x$ solution by taking the limit as $A \to 0$.
(Remark: this does include the sine solution you expected, since
$$ \sin{ax} = \frac{\sinh{iax}}{i}, $$
so taking $A=ia$ gives the sine solution.
Therefore we have shown

If $f(x)$ is thrice-differentiable and satisfies the functional equation
  $$ f(x)+f(y) = 2f\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right), $$
  then $f(x)$ is one of
  
  
*
  
*$0$
  
*$x$
  
*$A^{-1}\sinh{Ax}$, some $A \in \mathbb{C}$.
  

EDIT:
In fact, we only need twice-differentiability, because setting $x=0$,$y=2z$ in (3) gives
$$ 1 = (f'(z))^2-f(z)f''(z), $$
which is a thoroughly rotten-looking DE, but we can solve it by noticing that
$$ \left(\frac{f'}{f}\right)' = \frac{ff''-f'^2}{f^2}, $$
so an equivalent is
$$ -\frac{1}{f^2} = \left(\frac{f'}{f}\right)' $$
Multiply by $2f'/f$:
$$ -2\frac{f'}{f^3} = 2\left(\frac{f'}{f}\right)\left(\frac{f'}{f}\right)' $$
Integrating this once, we have
$$ a^2+\frac{1}{f^2} = \frac{f'^2}{f^2}, $$
and rearranging this gives
$$ 1 = \frac{f'^2}{a^2 f^2+1} $$
Taking the positive square root (since $f'(0)=1$) and integrating,
$$ z-0 = \int_0^{f(z)} \frac{dt}{\sqrt{1+a^2 t^2}} $$
Now, this is easy to integrate by setting $u=a^{-1}\sinh{t}$: then the integral just simplifies to $a^{-1}\int_0^{\arg\sinh{(af(z))}} du =\arg\sinh{(af(z))}, $
and hence $f(z)=\frac{1}{a}\sinh{az}$ as before. (And $a=0$ gives the $f(x)=x$ solution.)
