Proof Using Trigonometry

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We want to show it is always possible to solve

$\displaystyle A\cos(\omega t + \phi) = A_1\cos(\omega t + \phi_1) + A_2\cos(\omega t + \phi_2) + \cdots + A_N\cos(\omega t + \phi_N) \protect$

(A.2)

for

and $\phi$ , given $A_i, \phi_i$ for $i=1,\ldots,N$ . For each component sinusoid, we can write

$\displaystyle A_i\cos(\omega t + \phi_i)$	$\displaystyle =$	$\displaystyle A_i\cos(\omega t)\cos(\phi_i) - A_i\sin(\omega t)\sin(\phi_i)$
	$\displaystyle =$	$\displaystyle \left[A_i\cos(\phi_i)\right]\cos(\omega t) - \left[A_i\sin(\phi_i)\right]\sin(\omega t)$	(A.3)

Applying this expansion to Eq. $\,$ (A.2) yields

$\begin{eqnarray*} \left[A\cos(\phi)\right]\cos(\omega t) &-&\left[A\sin(\phi)\ri... ...a t) - \left[\sum_{i=1}^N A_i\sin(\phi_i)\right]\sin(\omega t). \end{eqnarray*}$

Equating coefficients gives

$\displaystyle A\cos(\phi)$	$\displaystyle =$	$\displaystyle \sum_{i=1}^N A_i\cos(\phi_i) \isdefs x$
$\displaystyle A\sin(\phi)$	$\displaystyle =$	$\displaystyle \sum_{i=1}^N A_i\sin(\phi_i) \isdefs y. \protect$	(A.4)

where

and

are known. We now have two equations in two unknowns which are readily solved by (1) squaring and adding both sides to eliminate $\phi$ , and (2) forming a ratio of both sides of Eq. $\,$ (A.4) to eliminate

. The results are