Trigonometric identities from differential equation If one knows that the solutions of $y''+ y=0$ are two functions $s(x)$ and $c(x)$, and we know that $s(0)=0$, $s'(0)=1$, $c(0)=1$, $c'(0)=0$, then how can one start to prove that $s(x+a)=s(x)c(a)+c(x)s(a)$? What is it that has to be observed in order for one to understand how to initiate the proof?
One is not supposed to use the actual trigonometric functions in this case. Here we assume that we know nothing about sin and cos.
 A: Here's a hint: You can use the Existence and Uniqueness theorem for linear differential equations to prove many different identities.
If you need to prove an identity $y_1(x)=y_2(x)$, first prove that both $y_1(x)$ and $y_2(x)$ are solutions to the provided differential equation. Then show that $y_1(a)=y_2(a)$ and $y_1'(a)=y_2'(a)$ for a value $a$ by using the initial conditions. Because there can only be one such solution, $y_1(x)$ must equal $y_2(x)$.
You'll want to use this approach to first prove $s'(x) = c(x)$ and $c'(x) = -s(x)$ which will allow you to differentiate each side of the identity when you need it. Then you can use the same approach to prove $s(x+a)=s(x)c(a)+c(x)s(a)$.
A: HINT:
Using this,
$$y=A\cos ax+B\sin ax$$ where $A,B$ are arbitrary constants to be determined by $$s(0)=c'(0)=0,s'(0)=c(0)=1$$
A: Not an answer but, by a series approach can you establish that
$$(x+a) - (x+a)^3/3! + ( x+a)^5/5!  - ...$$
$$ =( x - x^3/3! +  x^5/5!  -...) \cdot (  1 -  a ^2/2! + a^4/4! - ... ) +$$
$$(1- x^2/2! +x^4/4!- ... )\cdot ( a - a ^3/3! + a^5/5! - ....)  ? $$
A: We use the following Theorem: [Page 82 of  Simmons Differential Equations with Applications and Historical Notes]  If P(x), Q(x), and R(x) be continuous functions on a closed interval [a,b].  If x0 is any point in [a. b], and if y0 and y0' are any numbers whatever, then equation y"+P(x) y'+ Q(x)y=R(x) has one and only one solution y(x) on the entire interval such that y(x0)=y0, and y'(x0)=y0'.  
Y1=**s(x+a) is a solution of y"+y=0.
**Y2=**s(x)c(a)+c(x)s(a) is also a solution of y"+y=0.
Also, Y1(0)=s(a) and Y2(0)=s(a).  Y1'(0)=s'(x+a)|x=0=s'(0)=1
 and Y2'(0)=s'(x)c(a)+c'(x)s(a)|x=0  =  1.  **Hence by Theorem Y1=Y2.
