 # In Physics, the Simple Harmonic Oscillator is represented by the equation d^{2}x/dt^{2}=-\omage^{ Pam Stokes 2022-01-16 Answered

In Physics, the Simple Harmonic Oscillator is represented by the equation ${d}^{2}x/d{t}^{2}=-{\omega }^{2}x$.
By using the characteristic polynomial, you get solutions of the form $x\left(t\right)=A{e}^{i\omega t}+B{e}^{-i\omega t}$, I get that you Euler's formula ${e}^{i\theta }=\mathrm{cos}\theta +i\mathrm{sin}\theta$, but I can't seem to find my way all the way to the ''traditional form'' of $D\mathrm{cos}\omega t+C\mathrm{sin}\omega t$.

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However, if you assume the function x(t) is real, then they are related as $\stackrel{―}{A}=B$.
This can be seen since $\stackrel{―}{x\left(t\right)}=x\left(t\right)$ which gives us $\stackrel{―}{A}=B$ (since are orthonormal)
Rewriting the complex exponentials as sine and cosine we get
$x\left(t\right)=A\left(\mathrm{cos}\left(\omega t\right)+i\mathrm{sin}\left(\omega t\right)\right)+B\left(\mathrm{cos}\left(\omega t\right)-i\mathrm{sin}\left(\omega t\right)\right)=\left(A+B\right)\mathrm{cos}\left(\omega t\right)+i\left(A-B\right)\mathrm{sin}\left(\omega t\right)$
Now as proved before $\stackrel{―}{A}=B$ and hence $\left(A+B\right)$ is real and $\left(A-B\right)$ is imaginary and hence we can write where C, D are real number.
And the solution becomes
$x\left(t\right)=C\mathrm{cos}\left(\omega t\right)+D\mathrm{sin}\left(\omega t\right)$
where $C,D\in R$

###### Not exactly what you’re looking for? Robert Pina

Remember that $x\left(t\right)=A{e}^{i\omega t}+B{e}^{-i\omega t}+B{e}^{-i\omega t}$ is a solution for any values of A and B which means that choosing $A=B=\frac{1}{2}$ result in the solution
${x}_{1}\left(t\right)={e}^{i\omega t}+{e}^{-i\omega t}$
Euler's fomula, along with the properties , simplifies this solution to
${x}_{1}=\mathrm{cos}\left(\omega t\right)$
Similarly, choosing gives the solution ${x}_{2}\left(t\right)=\mathrm{sin}\left(\omega t\right)$
Finally, by linearity you can conclude that the linear combination $C{x}_{1}+D{x}_{2}$ is also a solution.

###### Not exactly what you’re looking for? alenahelenash
the answers given are absolutely fine as they show the way from ${e}^{±i\omega t}$ to sin and cos. It might be even shorter if you use that $\mathrm{sin}\left(x\right)=\frac{1}{2i}\left({e}^{ix}-{e}^{ix}\right)$ and $\mathrm{cos}\left(x\right)=\frac{1}{2}\left({e}^{ix}+{e}^{-ix}\right)$ This can be interpreted as a definition for the functions or seen from its series definitions.