Harmonic addition theorem

Question

In order to satisfy the following equality: $$a\cos{\theta}+b\sin{\theta}=r\cos{\left({x-x_0}\right)}$$

$x_0 = \tan^{-1}{\frac{b}{a}},$ and $a^2 + b^2 = r^2$. The latter statement implies $r = \pm\sqrt{a^2+b^2}$. (Harmonic addition theorem)

But how should you know whether to use the positive or negative $r$?

EDIT

Through testing, I was able to find that the sign of $r$ depends on the sign of $a$. But can someone help me understand why this is?

My apologies, my question was already answered here: https://math.stackexchange.com/q/4365486 — Mailbox, Apr 29 '22 at 13:50

score 0 · Answer 1 · edited Nov 17 '22 at 23:14

0

$a{\cos(x)} + b{\sin(x)} = r{\cos(x_0)\cos(x)} +r{\sin(x_0)\sin(x)}$ Thus, $a=r\cos(x_0) \implies r= \frac{a}{\cos(x_0)}$. Since you've already determined $x_0$, you can deduce what the sign of $\cos(x_0)$ is (for example by looking at which quarter of the unit circle the angle is). Since the sign of $a$ is known as the initial condition, now you know the sign of $r$, which is what you're looking for

edited Nov 17 '22 at 23:14

Mailbox

896

answered Apr 28 '22 at 18:03

Sgg8

1,362

Couldn’t you use the same logic for b? That expansion implies that $b = r\sin{x_0} \iff r = \frac{b}{\sin{x_0}}$. – Mailbox Apr 29 '22 at 13:08
@Mailbox yes, definitely. Choose whichever you want – Sgg8 Apr 29 '22 at 13:19

Harmonic addition theorem

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