if $\sin{(ax)}+\sin{(bx)}=\sin{(cx)}+\sin{(dx)}$,show $a=c$ or $a=d$ 
let $a,b,c,d>0$ and such for any real numbers $x\in R$ have
  $$\sin{(ax)}+\sin{(bx)}=\sin{(cx)}+\sin{(dx)}$$
  show that
  $a=c$ or $a=d$

I try it,becasue 
$$\sin{\left(\dfrac{(a+b)x}{2}\right)}\cos{\left(\dfrac{(a-b)x}{2}\right)}=\sin{\left(\dfrac{(c+d)x}{2}\right)}\cos{\left(\dfrac{(c-d)x}{2}\right)}$$
then I don't known how to continue

If you differentiate  and put $x=0$, you get $a+b=c+d$. Derivating two more time and putting again $x=0$, you get $a^3+b^3=c^3+d^3$. Simplifying by $a+b=c+d>0$ gives $a^2-ab+b^2=c^2-cd+d^2$, hence $(a+b)^2-3ab=(c+d)^2-3cd$ and $ab=cd$.  It is easy to finish. 
