Wikipedia:Reference desk/Archives/Mathematics/2022 January 12

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

January 12

The sign of harmonic addition theorem

 Resolved

Harmonic addition theorem has the following equations

$a\cos x+b\sin x=c\cos(x+\varphi )$

(1)

$c=\operatorname {sgn}(a){\sqrt {a^{2}+b^{2}}}$

(2)

given that $a\neq 0$. In addition, the following equation can be found in the citation

$a=c\cos \varphi$

(3)

which implies

$c={\frac {a}{\cos \varphi }}$

(4)

Now, consider the case

$a>0$

(5)

${\frac {\pi }{2}}<\varphi <\pi$

(6)

According to (5) and (2), $c$ should be positive. According to (5), (6) and (4), $c$ should be negative. Aforementioned results of $c$ seem to be inconsistent. So what mistakes have I made? - Justin545 (talk) 17:36, 12 January 2022 (UTC)

Equation (2) is accompanied by $\varphi =\arctan(-b/a)$. The function arctan is in principle multivalued, but its value is conventionally restricted to $-\pi /2<\varphi <\pi /2$. The sign of c comes from that convention. In your equation (6) you break that convention, and if you insist on breaking it, then you must switch the sign in (2). --Wrongfilter (talk) 18:17, 12 January 2022 (UTC)

Thanks you for the help. Indeed, signs of $a$ and $c$ are the same if $-\pi /2<\varphi <\pi /2$. And it seems signs of $a$ and $c$ are different if $\pi /2<\varphi <3\pi /2$. So, if I understand correctly, how about rewrite (2) as

$c=\operatorname {sgn}(\cos \varphi )\operatorname {sgn}(a){\sqrt {a^{2}+b^{2}}}$

(7)

in order to make sure (4) and (7) are consistent for all angles. - Justin545 (talk) 20:04, 12 January 2022 (UTC)

Or maybe replace $a$ in (2) by the RHS of (4):

$c=\operatorname {sgn} \left({\frac {a}{\cos \varphi }}\right){\sqrt {a^{2}+b^{2}}}$

(8)

given that $\cos \varphi \neq 0$ - Justin545 (talk) 20:13, 12 January 2022 (UTC)

I'm not sure what you're really trying to achieve and for some reason you always omit the second equation $\varphi =\arctan(-b/a)$ (or $\tan \varphi =-b/a$, if you prefer). You need two equations to find $(c,\varphi )$. --Wrongfilter (talk) 20:31, 12 January 2022 (UTC)

Okay, those are thoughtless replies. I am sorry for bothering you with the replies. - Justin545 (talk) 20:51, 12 January 2022 (UTC)