Harmonic Trig Form

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math

Is a way to express an equation in terms of \( \sin x \) and \( \cos x \) as a scaled value of an offset cosine or sine. That is to say solving \( a \cos x + b \sin x = R \cos (x \pm \alpha) \) or \( a \cos x + b \sin x = R \sin (x \pm \alpha) \).

Note: Because we can rewrite \( \cos \) in terms of \( \sin \) (and vice-versa) the choice of sine or cosine on the RHS is mostly irrelevant.

If you can choose which form to use [either \( R \cos \) or \( R \sin \)] then better to choose the versions which when expanded give the same signs for the corresponding terms as the original expression.

The harmonic form is preferable to the original because the stretch from \( \cos x \) and \( \sin x \) is obvious. Given an equation \( 13 \cos (x + 67.4^{\degree}) \) we know the maximum is 13 and the minimum is -13.

Example Harmonic Form Conversion

To demonstrate consider converting \( 4 \cos x + 3 \sin x \) into a harmonic trig function \( R \cos (x \pm \alpha) \). By the professor terror identity, this expands into

\begin{align*} 4 \cos x + 3 \sin x &= R \cos x \cos \alpha + R \sin x \sin \alpha \\

                  &= R ( \cos x \cos \alpha + \sin x \sin \alpha )

\end{align*}

In this form \( \cos x \) can only come from the \( 4 \cos x \) term on the left hand side (and similar for \( \sin x \)), so if we equate the two terms into separate equations and cancel out the shared \(\cos x\) (and \( \sin x \)) terms we get:

\begin{align*} 4 = R \cos \alpha \
3 = R \sin \alpha \end{align*}

We can divide these two equations by each other to cancel out \( R \) and get \( \tan \alpha = \frac{3}{4} \), which we can solve to get \( \alpha = \tan^{-1} \frac{3}{4} \approx 36.9^{\degree} \).

\( R \) is simply the positive square root of \( \sqrt{a_2 + b_2} \), which in this case is \( R = \sqrt{4^2 + 3^2} = 5 \).

Therefore \[ 4 \cos x + 4 \sin x = 5 \cos (x - 36.9^{\degree}) \]