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From the differential equation $$y'' + y = 0$$ we assume two solutions $s(x)$ and $c(x)$ and I am trying to show that under the conditions $s(0) = 0, s'(0) = 1, c(0) = 1, c'(0) = 0$, that $$s(2x) = 2s(x)c(x)$$ and $$c(2x) = c^2(x) - s^2(x)$$

The answer to Properties of sin(x) and cos(x) from definition as solution to differential equation y''=-y does not answer my question because the answer seems to assume that $$s(x + x) = s(x)c(x) + s(x)c(x)$$ follows from the fact that $$s(x + a) = s(x)c(a) + s(a)c(x)$$ but I don't see how this follows especially because the former does not satisfy the differential equation.

Is the following a valid proof for these identities ($s(2x)$ and $c(2x)$)?

Prior knowledge that $\cos^2(x) = \frac{1}{2}(1 + \cos(2x))$ looks promising and differentiating gives the result for $\sin(2x)$. To show that the identity holds for $c^2(x)$, I show that $y = c^2x$ satisfies the differential equation for $y = \frac{1}{2}(1+c^2(x))$. I then show that $s^2(x)$ also satisfies the DE and by existence and uniqueness $$c^2(x) = \frac{1}{2}(1 + \cos(2x))$$

Is this approach valid? How can I demonstrate this property without cheating?

Thanks in advance.

jaslibra
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1 Answers1

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The previous problem does not start with the assumption that $s(x+a)=s(x)c(a)+s(a)c(x)$. It just says that $s(x+a)$ satisfies the differential equation $y''+y=0$, with some different initial conditions. Since both $s(x)$ and $c(x)$ satisfy the differential equation, any linear combination of those satisfy the differential equation (though not the initial conditions). So if we write $s(x+a)$ as a linear combination of $s(x)$ and $c(x)$, like $$s(x+a)=f_1(a)s(x)+f_2(a)c(x)$$we need to find the functions $f_1$ and $f_2$. If we plug in $x=0$ we get $$s(a)=f_2(a)$$ If we take the derivative with respect to $x$ we have $$c(x+a)=f_1(a)c(x)-f_2(a)s(x)$$ Once again we plug in $x=0$, and we get $$c(a)=f_1(a)$$ Therefore $$s(x+a)=c(a)s(x)+s(a)c(x)$$

Andrei
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