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 January 5th, 2017, 10:54 AM #1 Newbie   Joined: Jan 2017 From: Irvine California Posts: 3 Thanks: 0 composite argument properties sin(a+b) = sin(a)cos(b) + cos(a)sin(b) Of course there are more for cos etc. I am wondering if any of you fine gentlemen or ladies have any information on the history of this property: I.E: who discovered it and perhaps a location of their notes? I am also in search of a concise proof; that search is possible through google though if you happen to know of a location for such a thing a url or search term is appreciated. -Buffalo
 January 5th, 2017, 11:08 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,770 Thanks: 1424
 January 5th, 2017, 02:56 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 472 Thanks: 262 Math Focus: Dynamical systems, analytic function theory, numerics I doubt anyone knows who discovered this and I'm sure it was known long before it was rigorously proved. The most concise proof (also what I believe to be the "best" proof) is the following: For any $x \in \mathbb{R}$, we have $e^{ix} = \cos x + i \sin x$. It follows that for $a,b \in \mathbb{R}$ we have $$\sin(a+b) = \Im (e^{i(a+b)}) = \Im(e^{ia}e^{ib}) = \Im ((\cos a + i \sin a)(\cos b + i \sin b)) = \cos a \sin b + \cos b \sin a$$. The proof for $\cos(a+b)$ follows similarly by taking the real part. The magic of this computation is its so obvious once you've seen it that you never have to memorize these identities again. In fact, similar computations replace almost all of the trig identities you had to memorize as an undergrad.
 January 7th, 2017, 10:56 PM #4 Newbie   Joined: Jan 2017 From: Irvine California Posts: 3 Thanks: 0 Thank you both for your responses. However, skeeter yours certainly proves the relationships, but it seems so contrived and what I am searching for and will benefit from is learning exactly how a mind could complicate a formula so simple to Code: (cos α − cos β)^2 + (sin α − sin β)^2 = (cos(α − β) − 1)^2 + (sin(α − β) − 0)^2 And, then abstractly fit such a thing into the Pythagorean theorem. Those mental leaps seem impractically astronomical. So maybe there is another proof that can show me how one could step by step reach such a beautifully built proof. SDK, your proof seems like it may bear fruit in that very regard. Unfortunately I lack the level of mathematical nomenclature knowledge to fully grasp what you have written. I am trying to use the powers of the internet to decode your proof, however, the nature of it still eludes me. Perhaps, and I am sure you are busy and could spend your time carrying on more important tasks, you could explain.
January 8th, 2017, 12:51 AM   #5
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Quote:
 Originally Posted by Buffalo SDK, your proof seems like it may bear fruit in that very regard. Unfortunately I lack the level of mathematical nomenclature knowledge to fully grasp what you have written.
$\Im$ is the imaginary part of a complex number. That's the mysterious looking bit. Here's SDK's proof in more detail.

The key is Euler's formula. If $x$ is any real number, then $e^{ix} = \cos x + i \sin x$. This is really an amazing fact. It's the key that relates trigonometry to the complex numbers and the complex exponential function.

Then if $a$ and $b$ are real numbers we have

$e^{i(a + b)} = \cos (a + b) + i \sin (a + b)$ -- Call this equation (1)

On the other hand from the rules for exponents (which are valid for complex numbers as well as real numbers, something that needs proof), we have

$e^{i(a + b)} = e^{ai + bi} = e^{ai} e^{bi}$

$= (\cos a + i \sin a)(\cos b + i \sin b)$ -- Euler's formula twice

$= \cos a \cos b - \sin a \sin b + i(\cos a \sin b + \sin a \cos b)$ -- Call this equation (2)

(The last line comes from multiplying out the two binomials in the prior line and applying the relation $i^2 = -1$).

Now equating the respective real and imaginary parts of equations (1) and (2) gives

$\cos (a + b) = \cos a \cos b - \sin a \sin b$

and

$\sin (a + b) = \cos a \sin b + \sin a \cos b$

and as SDK notes, you never have to memorize these identities again. You just run through this little calculation whenever you need them.

<MathEdRant>
For this and many other reasons, high school trig should be taught based on the complex numbers, making life so much simpler. We don't use trig to measure the fields we're farming like we did a thousand years ago. Trig is not about triangles, it's about the unit circle. We care about $\sin$ and $\cos$ because they're the building blocks of Fourier series and modern digital signal processing. Math education needs to catch up.
</MathEdRant>

Last edited by Maschke; January 8th, 2017 at 01:02 AM.

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