-2

let $f,g$ be contiunous, bounded on $\Bbb R$. Suppose that $$f(x+y)=f(x)f(y)-g(x)g(y),\quad g(x+y)=f(x)g(y)+f(y)g(x),\ \forall\ x,y\in\ \Bbb R.$$ $f(0)=1$, $g(0)=0$. Show that for some $a\in\Bbb R$, $f(x)=\cos ax, g(x)=\pm \sin a x$.

xldd
  • 3,407
  • Just plug in the relation for sum of angles, https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Sine.2C_cosine.2C_and_tangent_of_multiple_angles – Gregory May 09 '17 at 00:01

1 Answers1

-1

It may be possible to prove from the conditions that $f$ and $g$ are derivable/differentiable, and from that find a differential equation for $f$ (and another for $g$), for which you know that the solution is $\sin$ or $\cos$.

[Sorry, I didn't try myself. If I'm wrong I'll delete it. :-( ]

Pablo H
  • 552