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Harmonic Addition Theorem

To convert an equation of the form

\begin{displaymath}
f(\theta)=a\cos\theta+b\sin\theta
\end{displaymath} (1)

to the form
\begin{displaymath}
f(\theta)=c\cos(\theta+\delta),
\end{displaymath} (2)

expand (2) using the trigonometric addition formulas to obtain
\begin{displaymath}
f(\theta) = c\cos\theta\cos\delta-c\sin\theta\sin\delta.
\end{displaymath} (3)

Now equate the Coefficients of (1) and (3)
$\displaystyle a$ $\textstyle =$ $\displaystyle c\cos\delta$ (4)
$\displaystyle b$ $\textstyle =$ $\displaystyle -c\sin\delta,$ (5)

so
\begin{displaymath}
\tan\delta = - {b\over a}
\end{displaymath} (6)


\begin{displaymath}
a^2+b^2 = c^2,
\end{displaymath} (7)

and we have
$\displaystyle \delta$ $\textstyle =$ $\displaystyle \tan^{-1}\left({- {b\over a}}\right)$ (8)
$\displaystyle c$ $\textstyle =$ $\displaystyle \sqrt{a^2+b^2}.$ (9)

Given two general sinusoidal functions with frequency $\omega$:
$\displaystyle \psi_1$ $\textstyle =$ $\displaystyle A_1\sin(\omega t+\delta_1)$ (10)
$\displaystyle \psi_2$ $\textstyle =$ $\displaystyle A_2\sin(\omega t+\delta_2),$ (11)

their sum $\psi$ can be expressed as a sinusoidal function with frequency $\omega$


$\displaystyle \psi$ $\textstyle \equiv$ $\displaystyle \psi_1+\psi_2= A_1[\sin(\omega t)\cos\delta_1+\sin\delta_1\cos(\omega t)]+A_2[\sin(\omega t)\cos\delta_2+\sin\delta_2\cos(\omega t)]$  
  $\textstyle =$ $\displaystyle [A_1\cos\delta_1+A_2\cos\delta_2]\sin(\omega t)+[A_1\sin\delta_1+A_2\sin\delta_2]\cos(\omega t).$ (12)

Now, define
\begin{displaymath}
A\cos\delta \equiv A_1\cos\delta_1+A_2\cos\delta_2
\end{displaymath} (13)


\begin{displaymath}
A\sin\delta \equiv A_1\sin\delta_1+A_2\sin\delta_2.
\end{displaymath} (14)

Then (12) becomes
\begin{displaymath}
A\cos\delta \sin(\omega t)+A\sin\delta\cos(\omega t) = A\sin(\omega t+\delta).
\end{displaymath} (15)

Square and add (13) and (14)
\begin{displaymath}
A_2 = {A_1}^2+{A_2}^2+2A_1A_2\cos(\delta_2-\delta_1).
\end{displaymath} (16)

Also, divide (14) by (13)
\begin{displaymath}
\tan\delta = {A_1\sin\delta_1+A_2\sin\delta_2 \over A_1\cos\delta_1+A_2\cos\delta_2},
\end{displaymath} (17)

so
\begin{displaymath}
\psi = A\sin(\omega t+\delta),
\end{displaymath} (18)

where $A$ and $\delta$ are defined by (16) and (17).


This procedure can be generalized to a sum of $n$ harmonic waves, giving

\begin{displaymath}
\psi = \sum_{i=1}^n A_i\cos (\omega t+\delta_i)= A\cos (\omega t+\delta),
\end{displaymath} (19)

where
$\displaystyle A^2$ $\textstyle \equiv$ $\displaystyle \sum_{i=1}^n\sum_{j=1}^n A_iA_j\cos(\delta_i-\delta_j)$ (20)
  $\textstyle =$ $\displaystyle \sum_{i=1}^n {A_i}^2 + 2\sum_{i=1}^n \sum_{j>i}^n A_iA_j\cos(\delta_i-\delta_j)$ (21)

and
\begin{displaymath}
\tan\delta = {\sum_{i=1}^n A_i\sin \delta_i\over \sum_{i=1}^n A_i\cos \delta_i}.
\end{displaymath} (22)



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© 1996-9 Eric W. Weisstein
1999-05-25