Proof Using Trigonometry

We want to show it is always possible to solve

$$A\cos(\omega t + \phi) = A_1\cos(\omega t + \phi_1) + A_2\cos(\omega t + \phi_2) + \cdots + A_N\cos(\omega t + \phi_N) \protect$$ (A.2)

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

$$A_i\cos(\omega t + \phi_i) = A_i\cos(\omega t)\cos(\phi_i) - A_i\sin(\omega t)\sin(\phi_i)$$

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

Applying this expansion to Eq.(A.2) yields

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

Equating coefficients gives

$$A\cos(\phi) = \sum_{i=1}^N A_i\cos(\phi_i) \isdefs x$$

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

where $x$ and $y$ 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 $A$ . This gives

$$\begin{eqnarray*} A &=& \pm\sqrt{x^2+y^2}\\ \phi &=& \tan^{-1}\left(\frac{y}{x}\right) \end{eqnarray*}$$

for any values of $A_i$ and $\phi_i$ . Since $\tan^{-1}(y/x) = \tan^{-1}(-y/-x)$ , we have $\phi\in[-\pi/2,\pi/2)$ . To impose $A\ge 0$ and $\phi\in[-\pi,\pi)$ , a four-quadrant arctangent$(y,x)$ must be used, normally written atan2(y,x) in computer languages.