Proof Using Trigonometry

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 $ A$ 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

\left[A\cos(\phi)\right]\cos(\omega t)
...a t)
- \left[\sum_{i=1}^N A_i\sin(\phi_i)\right]\sin(\omega t).

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 $ 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$. The results are

A &=& \sqrt{x^2+y^2}\\
\phi &=& \tan^{-1}\left(\frac{y}{x}\right)

which has a unique solution for any values of $ A_i$ and $ \phi_i$.

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Half-Angle Tangent Identities