# Sum of $n$ Out-of-phase Sinusoids with the Same Frequency

\begin{aligned} \sum\limits_{n}e^{i(x+\varphi_n)}&=\sum\limits_{n}(\cos(x+\varphi_n)+i\sin(x+\varphi_n))\\ &=\sum\limits_{n}\cos(x+\varphi_n)+\sum\limits_{n}i\sin(x+\varphi_n)\\ &=\sum\limits_{n}\cos(x+\varphi_n)+i\sum\limits_{n}\sin(x+\varphi_n)\\ \end{aligned}

\begin{aligned} \sum\limits_{n}e^{i(x+\varphi_n)}&=\sum\limits_{n}e^{ix+i\varphi_n}\\ &=\sum\limits_{n}e^{ix}e^{i\varphi_n}\\ &=e^{ix}\sum\limits_{n}e^{i\varphi_n}\\ &=e^{ix}\sum\limits_{n}(\cos(\varphi_n)+i\sin(\varphi_n))\\ &=e^{ix}\left(\sum\limits_{n}\cos(\varphi_n)+\sum\limits_{n}i\sin(\varphi_n)\right)\\ &=e^{ix}\left(\sum\limits_{n}\cos(\varphi_n)+i\sum\limits_{n}\sin(\varphi_n)\right)\\ &=e^{ix}(A+iB)\\ &=e^{ix}\sqrt{A^2+B^2}e^{i\arg(A+iB)}\\ &=\sqrt{A^2+B^2}e^{ix+i\arg(A+iB)}\\ &=\sqrt{A^2+B^2}e^{i(x+\arg(A+iB))}\\ &=\sqrt{A^2+B^2}(\cos(x+\arg(A+iB))+i\sin(x+\arg(A+iB)))\\ &=\sqrt{A^2+B^2}\cos(x+\arg(A+iB))+i\sqrt{A^2+B^2}\sin(x+\arg(A+iB)) \end{aligned}

$\boxed{ \begin{cases} \sum\limits_{n}\sin(x+\varphi_n)=\sqrt{A^2+B^2}\sin(x+\arg(A+iB))\\ \sum\limits_{n}\cos(x+\varphi_n)=\sqrt{A^2+B^2}\cos(x+\arg(A+iB))\\ A=\sum\limits_{n}\cos(\varphi_n)\\ B=\sum\limits_{n}\sin(\varphi_n) \end{cases}}$

Example:

\begin{aligned} &\phantom{=}\sin(2\pi(t+e))+\sin(2\pi t+69)+\sin(2\pi t+42^\circ)\\ &=\sin(2\pi t+2\pi e)+\sin(2\pi t+69)+\sin(2\pi t+42^\circ)\\ &=\sqrt{A^2+B^2}\sin(2\pi t+\arg(A+iB)) \end{aligned} $\begin{cases} A=\cos(2\pi e)+\cos(69)+\cos(42^\circ)\approx1.53856064504\\ B=\sin(2\pi e)+\sin(69)+\sin(42^\circ)\approx-0.425861365902 \end{cases}$ $\sqrt{A^2+B^2}\approx1.59641058674$ $\arg(A+iB)=\tan^{-1}\left(\frac BA\right)\approx-15.4716657207^\circ$ $\sin(2\pi(t+e))+\sin(2\pi t+69)+\sin(2\pi t+42^\circ)\approx\boxed{1.60\sin(2\pi t-15.5^\circ)}$

Note by Gandoff Tan
9 months, 3 weeks ago

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