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Does anyone have any insight into the trig sum and difference formulas? The formulas in themselves are very elegant, but I don't really like the proofs that have been given, even the geometric proofs. I feel that none of them address the following points:

cos(A+B) = cosAcosB - sinAsinB

  • If you keep one angle constant in this formula, say angle A, and change angle B, why will it trace out the exact cosine curve, just displaced by angle A? This doesn't seem obvious at all from the formula.

  • Also, the geometric proof is nice, but it doesn't show (to me) why adding angle A to B is the same as adding angle B to A

  • Earlier today I asked if trigonometry could exist in one dimension, and I think the answer is yes

  • In another question on this site, someone brought up the idea of matrix multiplication - now I know the mechanics of it, but have literally no idea why it works

So, if anyone had any extra insight into this weird formula, that would really be greatly appreciated!! I realise that this question may seem very strange, maybe even stupid, but hopefully you see where I am sort of coming from. Thanks.

I have been scouring the net for quite a while now but haven't gained much intuition. I do have a sort of obsession with really trying to understand all the formulas that they teach me at high school (final year coming up), and in a way it has delayed my progression in the subject.

So I guess another question would be - is it even possible to gain a deep intuition into high school mathematics, so for example, to feel as natural manipulating these trig equations as multiplication, or manipulating logarithms for example.

Kevin
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  • In Mathematics you never understand things, you just get used to them –  Oct 26 '14 at 11:51
  • @DigitalBrain: That is not true. I can affirm that I fully understand almost everything that I have learnt so far, and I am fully clear of exactly what I still don't understand. You can read my answer to this question in particular and see if what I say about these trigonometric identities makes sense. However, I'm afraid that you would need to know a fair bit of calculus to fully understand everything I say. – user21820 Oct 26 '14 at 12:00
  • By the way, your question is not strange nor stupid. In the past I had exactly the same questions and only years later did I find the answers that are fully satisfactory to myself. And I doubt it will actually delay your progression because you won't lose any of what you learn in contrast to if you "just blindly memorized facts and techniques". If you find something too hard to understand or figure out at some point, you can always put it aside temporarily and eventually come back to it when you think you have the necessary skills to tackle it. – user21820 Oct 26 '14 at 12:11
  • Kevin, have you studied or seen vectors in your mathematical studies? – Ian Mateus Oct 26 '14 at 13:07
  • I'm not sure if this will give the deep understanding that you're seeking, but have a look at this geometric proof. http://math.stackexchange.com/questions/910590/proof-of-trigonometric-identity-sinab-sin-a-cos-b-cos-a-sin-b/911092#911092 – John Joy Oct 26 '14 at 16:07
  • Thanks @JohnJoy that is a really nice proof – Kevin Oct 27 '14 at 08:29
  • @IanMateus I have a really basic knowledge from school, but have been looking around in my own spare time, trying to figure out the dot-product (I don't think I've gotten it yet, just don't fully understand why the dot product remains invariant under a rotation of the coordinates) – Kevin Oct 27 '14 at 08:31
  • @Kevin Check out http://math.oregonstate.edu/bridge/mathml/dot+cross.xhtml. The author claims that the problem is that dot and cross products are taught a-- backwards. They are usually defined as algebraic operations, whereas the website proposes that it would be better to start with a geometric definition and then derive algebraic consequences. – John Joy Oct 27 '14 at 13:26
  • @JohnJoy: Nice proof! For real angles that is. Incidentally, it comes naturally from considering angles as 2d rotations, which can be represented as 2*2 matrices. When you multiply out the two matrices for two angles, you get the sum rules for exactly the same reason your answer gets it. – user21820 Oct 27 '14 at 15:34
  • @Kevin: As for dot product of vectors $u,v$, at your level you can assume the vectors as being in $\mathbb{R}^n$ for some $n \in \mathbb{N}$ and consider them as $n \times 1$ (column) matrices, in which case $u \cdot v = u^T v$. A rotation matrix $R$ has the property that $R^T R = I$, because the length of each column of $R$ is $1$ and any two distinct columns of $R$ are orthogonal. Thus you get $(Ru) \cdot (Rv) = (Ru)^T Rv = u^T R^T R v = u^T v$. This is a purely formal derivation but you can 'see the result coming' once you realise that $R^T R = I$. – user21820 Oct 27 '14 at 15:46
  • To explain $R^T R = I$ in detail, $R$ is supposed to take the elementary vectors to orthogonal unit length vectors, so each column of $R$ must be a unit vector, and the dot product of any two distinct columns is $0$. Note that we can define rotation matrices as those $R$ that satisfy $R^T R = I$, but it would allow reflections to be rotations, as well as compositions of rotations and reflections. As proven above, they all preserve the dot product. Some people require rotation matrices $R$ to also satisfy $det(R) = 1$, and call rest (which must satisfy $det(R) = -1$) improper. – user21820 Oct 27 '14 at 15:57

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I believe you will only be satisfied when you understand $\cos$ and $\sin$ as related to $\exp$ via:

$\cos(z) = \frac{1}{2} ( e^{iz} + e^{-iz} )$ for any $z \in \mathbb{C}$

$\sin(z) = \frac{1}{2i} ( e^{iz} - e^{-iz} )$ for any $z \in \mathbb{C}$

I still think that the best way to define them all is via their power series which arises naturally from solving the first and second order ordinary linear differential equations, which in turn arise from naturally occurring phenomena. Of course, to do so would require various concepts such as limits and differentiability, but it is in my opinion very well motivated. After that, you can proceed as in https://math.stackexchange.com/a/802678/21820, which gives all the fundamental properties, and periodicity especially arises as the path of $\exp(it)$ along the unit circle for real $t$.

Now with these it is immediately clear that the special characteristics of all the common trigonometric identities arise from the characteristics of the exponential function, in particular that $\exp(x+y) = \exp(x) \exp(y)$. You can try proving all of them by just converting them to the equivalent identities for the exponential function. For example here is the particular formula you asked about:

$\cos(a+b) = \frac{1}{2} ( e^{i(a+b)} + e^{-i(a+b)} ) = \frac{1}{2} ( e^{ia} e^{ib} + e^{-ia} e^{-ib} )$

$\cos(a) \cos(b) - \sin(a) \sin(b) = \frac{1}{4} ( e^{ia} + e^{-ia} ) ( e^{ib} + e^{-ib} ) + \frac{1}{4} ( e^{ia} - e^{-ia} ) ( e^{ib} - e^{-ib} )$

$ = \frac{1}{2} ( e^{ia} e^{ib} + e^{-ia} e^{-ib} )$

And if you only need real $a,b$:

$\cos(a+b) = Re(e^{i(a+b)}) = Re(e^{ia}e^{ib}) = Re(e^{ia}) Re(e^{ib}) - Im(e^{ia}) Im(e^{ib})$

$ = \cos(a) \cos(b) - \sin(a) \sin(b)$ where the third equality can be checked directly from the multiplication of complex numbers.

Concerning your side question about matrix multiplication, left-multiplication of a matrix with a vector corresponds to some linear transformation of a vector. To get the desired property that the product of two matrices corresponds to the (non-commutative) composition of the corresponding linear transformations, we necessarily have to define the matrix multiplication exactly the way you have been taught. You should derive it for yourself to see that it is indeed the only way.

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user21820
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    Nice! but see to solve one complication you're creating another one, a bit un-$\mathbb R$ealistic. –  Oct 26 '14 at 12:13
  • @DigitalBrain: It may seem unrealistic at first but when you realize all the wonderful properties of $\mathbb{C}$ it will seem completely natural. For instance it is algebraically closed, and analytic functions have extremely nice behaviour. These are crucial when you want to solve the differential equations in the first place, because you need complex numbers for some differential equations such as simple harmonic motion. Analyticity of the exponential function is essentially built in to the definition by the differential equation. – user21820 Oct 26 '14 at 12:23
  • So, you got used to them? –  Oct 26 '14 at 12:32
  • @DigitalBrain: No I understood why it had to be that way. It is the natural result of wanting solutions to $x^2+1=0$ and it is easy to prove that adding $i$ defined to satisfy the equation creates a field, and the rest follows from calculus. Of course there are plenty of mathematical areas which I still don't understand, but these fundamental constructions are not one of them anymore. – user21820 Oct 26 '14 at 12:35
  • Thanks @user21820, this provides a clear picture of how complex numbers are linked to exponentials - I spent math class this lesson reading up on Euler's formula, and now I mostly get it! Complex multiplication is still wishy washy in my head, but yeah thanks, this was wonderful!! – Kevin Oct 27 '14 at 10:03
  • @Kevin: You're welcome! Complex multiplication can be viewed as a scale followed by a rotation. This is succinctly represented by the form $r e^{it}$ for some $r,t \in \mathbb{R}$, where $r$ is the scale and $t$ is the rotation angle. The form $x+yi$ for some $x,y \in \mathbb{R}$ is useful when you need to view the complex number as a vector. – user21820 Oct 27 '14 at 15:22
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When you are at a certain mathematical level you have at your disposal and can use all the material that has been presented to you up to this point. You will observe some patterns, like the addition formulas for the trig functions. In many cases you have a full understanding of what's going on, and in other cases you can parse the proof of such patterns step for step, but you feel uneasy.

Another example: You understand the construction for bisecting an angle, and you hope that the teacher will show you a method for trisecting an angle next week. But, as we all know, there is no such method. The insight why this is so only comes after you have climbed to a higher level of mathematical knowledge and understanding.

Same thing with the trigonometric functions: That they behave as they do is a miracle to you now, but it can be understood from a (moderately) higher mathematical viewpoint. Once you have reached this point it will be obvious to you that all the formulas (and other properties) valid for the trig functions stem from one basic fact: the addition theorem ("law of exponents") for the exponential function, valid not only for real, but also for complex exponents: $$e^{z_1+z_2}=e^{z_1}\cdot e^{z_2}\ .$$

Christian Blatter
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  • I agree that in some cases it is impossible to understand why something is true at the current level, such as why trisecting a general angle using straightedge and compass in finitely many steps is impossible, but in general many mathematics educators introduce a lot of arbitrary facts without any justification even if it is not hard to provide justification. Also, I much dislike the way that mathematical theorems are often presented without first giving full motivation to why we want to consider such a theorem in the first place, nor how we can derive the proof, not just the theorem. – user21820 Oct 27 '14 at 15:24
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    So personally I believe that if the proof is within the level of the student, the proof itself needs to be fully motivated. If it cannot be, in other words some step just 'fell from the sky', it indicates that the teacher does not yet understand the theorem fully! I'm sure that most high-school teachers do not realize that the properties of trigonometric functions are actually arising from the properties of the complex exponential function as we've both said in our answers, in which case the only answer they can give the student is "The proof works.", which would be hardly satisfying. – user21820 Oct 27 '14 at 15:29