A Complex Proof of the Harmonic Addition Theorem

[@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!

[@ducovanw9120]

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

[@DrBarker]

Thank you!

[@youssef8955]

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

[@Jack_Callcott_AU]

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

[@lambda5949]

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).

[@seanfraser3125]

Yes, the polar form of a complex number completely describes the number. In fact, scalar magnitude and direction of vectors is a great analogy: the modulus is the magnitude, and e^(i*theta) where theta is the argument is the direction. Notice that the modulus of e^(i*theta) for any real theta is always 1, so it gives only a direction.

[@gavintillman1884]

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

[@squeezy8414]

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

[@DrBarker]

Yeah, I think the most straightforward way to prove this would be to start with Rcos(θ+α), then show that R and α have the desired values. But I do like proofs where we start from Acos(θ) + Bsin(θ) and work in the other direction, because this is the direction in which I would usually apply this result (to simplify a sum of cos and sin functions), and it hopefully gives slightly more of a sense of where the result comes from.

[@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)) = Re^(i(a+t))=Rcos(a+t)+iRsin(a+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).

[@aboodahmad8236]

We use a lot of COMPLEX stuff:)

[@matematicasemplice]

Bravo