Multiple of Sine plus Multiple of Cosine/Sine Form
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Theorem
- $p \sin x + q \cos x = \sqrt {p^2 + q^2} \map \sin {x + \arctan \dfrac q p}$
Proof
Let it be assumed that $p \sin x + q \cos x$ can be expressed in the form $M \map \sin {x + \phi}$.
Then:
\(\ds p \sin x + q \cos x\) | \(=\) | \(\ds M \map \sin {x + \phi}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac p M \sin x + \frac q M \cos x\) | \(=\) | \(\ds \map \sin {x + \phi}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin x \cos \phi + \cos x \sin \phi\) | Sine of Sum | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sin x \cos \phi\) | \(=\) | \(\ds \frac p M \sin x\) | equating terms in $\sin x$ | ||||||||||
\(\ds \cos x \sin \phi\) | \(=\) | \(\ds \frac q M \cos x\) | and $\cos x$ | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \sin \phi\) | \(=\) | \(\ds \frac q M\) | ||||||||||
\(\text {(2)}: \quad\) | \(\ds \cos \phi\) | \(=\) | \(\ds \frac p M\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sin^2 \phi + \cos^2 \phi\) | \(=\) | \(\ds \paren {\frac q M}^2 + \paren {\frac p M}^2\) | squaring and adding $(1)$ and $(2)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(=\) | \(\ds \frac {p^2 + q^2} {M^2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sqrt {p^2 + q^2}\) | \(=\) | \(\ds M\) |
Then:
\(\ds \tan \phi\) | \(=\) | \(\ds \frac {\sin \phi} {\cos \phi}\) | Tangent is Sine divided by Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {q / M} {p / M}\) | from $(1)$ and $(2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac q p\) | simplifying |
Hence the result.
$\blacksquare$