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

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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)