A Complex Proof of the Harmonic Addition Theorem

We look at a proof of the "harmonic addition theorem", using complex numbers, which allows us to write sums of the form Acos(θ) + Bsin(θ) as a single function Rcos(θ+α), for certain R and α.

00:00 Intro
00:28 Proof

[kindreon]

I was ready for a huge mess of trig identities, but the Euler's formula expressions made the proof very clean. It feels like a fun pattern could arise from recursive application like cos(t) + sin(t) = Rcos(t') then Rcos(t') + Rsin(t') = R'cos(t'') and so on. Thanks for sharing!

[youssef]

That was the nicest and the easiest approach I've seen so far!

[ducovanw]

I had heard about this proof but never seen it. Thank you for showing it!! Wonderful video as always!

[Jack_Callcott]

Very neat. It's something we need to know. Thanks.

[gavintillman]

Don’t think I ever saw this proven before. Nice.

[dirichlettt]

I did this by letting z=A-Bi such that R=|z|=sqrt(A^2+B^2) and a=arg(z), tan(a)= -B/A. The polar form of z is z = Re^(ia). Multiply both sides by w=e^(it) = cos(t)+isin(t) such that the rectangular forms are on one side of the equation and the polar forms are on the other: (A-Bi)(cos(t)+isin(t)) = Expanding the left side out and comparing real and imaginary components, we get two identities for the price of one: Acos(t)+Bsin(t)=Rcos(a+t) and Asin(t)-Bcos(t)=Rsin(a+t).

[lambda]

Awesome vid! Still a newbie to working with complex numbers, so I was just wondering: Is the product of the modulus and argument a complete description of a complex number (similar to the scalar magnitude and direction vectors), or was there some other process used to equate the (A/2 - Bi/2) and (R/2)*e^ia? (I'm unaccustomed to working with Euler's formulation of complex numbers, so sorry if my vector comparison is nonsensical).

[squeezy]

cool proof, is the method itself (i.e. expanding out cos(x+alpha) and then multiplying through by R and then comparing coefficients) considered a proof? It seems very intuitive to me and I can’t think of a particular case it wouldn’t work

[gunhasirac]

wow this proof is indeed very complex, but is not as complex as it might sound.