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I read in this post that one way to define sine and cosine is in terms of the differential equation $$y''=-y$$ along with some "initial conditions". Adding the conditions $y'(0)=1$ and $y(0)=0$ should suffice, since if $y_1$ and $y_2$ were two functions that complied with these properties then

$$(y_1'y_2-y_1y_2')'=y_1''y_2+y_1'y_2'-y_1'y_2'-y_1y_2''=y_1''y_2-y_1y_2''=y_1y_2-y_1y_2=0.$$ So there is some constant $c$ for which $y_1'y_2-y_1y_2'=c$, plugging in $0$ we get $$1(0)-0(1)=c=0.$$ Therefore $$\Big( \frac{y_1}{y_2}\Big) '=\frac{y_1'y_2-y_1y_2'}{y_1^2}=0,$$ forcing $y_1=ky_2$ and plugging in $0$ gives $1=k$.

I still need to show that one such function does exist, but I have no idea on how to do this (without appealing to another definition of sine and cosine). I would appreciate any help/thoughts.

Edit: Completed the proof (I hope) that $y_1=y_2$ by adding the condition $y(0)=0$, as suggested by Peter Foreman.

Sam
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  • @PeterForeman I'm guessing you mean $\text{sin}$ instead of $\text{cos}$? And even then, doesn't the condition $y'(0)=1$ force $A=1$? – Sam Jun 04 '20 at 20:01
  • Sorry I meant $$y(x)=\sin{(x)}+A\cos{(x)}$$ – Peter Foreman Jun 04 '20 at 20:05
  • @PeterForeman, you are right, but at the same time I can't seem to find the mistake in my proof that $y_1=y_2$, can you? Also, adding the condition $y(0)=0$ would fix the issue, right? – Sam Jun 04 '20 at 20:12
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    Your mistake is that you have only shown that $y_2(0)-y_1(0)=c$ i.e. when $x=0$ (it doesn't extend to the entire domain). But this is true because a general solution has $y(0)=A$ arbitrary as shown above. – Peter Foreman Jun 04 '20 at 20:17

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