Multiple of Sine plus Multiple of Cosine/Cosine Form

Theorem

$p \sin x + q \cos x = \sqrt {p^2 + q^2} \map \cos {x + \arctan \dfrac {-p} q}$

Proof

Let it be assumed that $p \sin x + q \cos x$ can be expressed in the form $M \map \cos {x + \phi}$.

Then:

\(\ds p \sin x + q \cos x\)

\(=\)

\(\ds M \map \cos {x + \phi}\)

\(\ds \leadsto \ \ \)

\(\ds \frac p M \sin x + \frac q M \cos x\)

\(=\)

\(\ds \map \cos {x + \phi}\)

\(\ds \)

\(=\)

\(\ds \cos x \cos \phi - \sin x \sin \phi\)

Cosine of Sum

\(\ds \leadsto \ \ \)

\(\ds \sin x \sin \phi\)

\(=\)

\(\ds \frac {-p} M \sin x\)

equating terms in $\sin x$

\(\ds \cos x \cos \phi\)

\(=\)

\(\ds \frac q M \cos x\)

and $\cos x$

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \sin \phi\)

\(=\)

\(\ds \frac {-p} M\)

\(\text {(2)}: \quad\)

\(\ds \cos \phi\)

\(=\)

\(\ds \frac q M\)

\(\ds \leadsto \ \ \)

\(\ds \sin^2 \phi + \cos^2 \phi\)

\(=\)

\(\ds \paren {\frac {-p} M}^2 + \paren {\frac q M}^2\)

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

\(\ds \)

\(=\)

\(\ds \frac {-p / M} {q / M}\)

from $(1)$ and $(2)$

\(\ds \)

\(=\)

\(\ds \frac {-p} q\)

simplifying

Hence the result.

$\blacksquare$