Memorizing ... A course of trigonometry can be surprisingly confusing and somewhat counter-mathematical, an increasing number of identities that seem to be unending sometimes, question is how can I understand these formulas intuitively? The way a mathematician does, not based on artificial definitions and symbols, but rather getting into the heart, the essence of the subject, a way of thinking that can make them trivial, give the feeling of why they work, aside from the way we prove them. For instance $$\cos \theta +\cos \varphi =2\cos \left({\frac {\theta +\varphi }{2}}\right)\cos \left({\frac {\theta -\varphi }{2}}\right)$$ Can lead to several questions, why did we multiply by a factor of $2$, why did we take the average of $\theta$ and $\varphi$, of course the way we chose to define $\cos(x)$ is what led to this fuzzy discovery, but how ? I believe, with no doubt that there should be a solution, as Poincaré put it : "Mathematics is the art of giving the same name to different things." Relating these identities with other concepts in Math, (Linear transformations and rotations in the case of Matrix form for the sum and difference formulae)as an example. Any help would be appreciated. Thanks for your time.


(1). Background:

Let $E$ be the Euclidean plane.

(i). If $f:E\to E$ is an isometry then $f$ is a bijection, so $f^{-1}:E\to E$ is also an isometry.

(ii).If an isometry $f:E\to E$ has 3 non-co-linear fixed points then $f=id_E.$

(iii). If isometries $f,g$ on $E$ agree on 3 non-co-linear points then $f=g$ because the isometry $g^{-1}f$ fixes those 3 points, so $g^{-1}f=id_E.$

(2). Choose an origin and orthogonal co-ordinate axes for $E.$

Any $2x2$ matrix $M(a)$ with top row $(\cos a, \;-\sin a)$ and bottom row $(\sin a,\;\cos a)$ can be regarded as a function on $E$ that maps $v=\binom {x}{y}\in E$ to $M(a)v\in E.$ The theorem of Pythagoras implies that $M(a)$ is an isometry.

The isometry $R(a)$ rotates each point of $E$ thru the angle $a$ counter-clockwise about $\binom {0}{0}.$ Note that $R(a)R(b)=R(a+b).$

Since $M(a)(v)= R(a)(v)$ when $v \in \{ \binom {0}{0},\binom {1}{0}, \binom {0}{1}\},$ we have $M(a)=R(a).$

$$\text {Therefore }\quad M(a)M(b)=R(a)R(b)=R(a+b)=M(a+b).$$ Comparing entries in the matrices $M(a)M(b)$ and $M(a+b),$ we obtain the trig angle-sum formulae as a necessary consequence of Part (1),,(i),(ii),(iii).


Here is an example where some light can be shed on another important trigonometric formula:


Let us give the names $f : x \to \cos^2(x)$ and $g : x \to \cos(2x)$.

  • $1$st approach: we already know the RHS of (1).

$f$ and $g$ share two properties: they are even and periodic with period $\pi$.

and it is evident that the graphical representation of $f$ is a shifted and reduced version of the graphical representation of $g$ (see figure).

  • $2$nd (heuristic) approach, where the RHS is unknown. We need here the concept of Fourier expansions (have you already had a lecture on this important subject ?).

$f$ being even and periodic with period $\pi$, possesses a Fourier expansion $\sum_{k=0}^{\infty} a_k \cos(kx)$.

A look at the graphical representation of $f$ shows that a good candidate for $a_0$ (quadratic mean value) is $1/2$;

Besides, the only frequency present in the spectrum of $f$ is twice the basic initial frequency (explaining the $2x$ instead of $x$). Thus there is a second term $a_2 \cos(2x)$. Because the maximum value of the RHS is 1, we have necessarily $a_2=1/2$. Tha'st all.

In fact (1) can be seen as a finite Fourier expansion.

Remark, one can, in almost the same manner, anticipate formulas of the same kind for any power $cos^n(x)$ as a Fourier decomposition.

enter image description here

  • 1
    $\begingroup$ I need a lecture on fourier analysis. $\endgroup$ – Simply Beautiful Art Jan 10 '17 at 23:32

If you recall Euler's formula

$$e^{ix}=\cos x+i\sin x$$

Then you will be able to basically derive all trig identities (using exponential function rules.

  • $\begingroup$ I don't want to derive/prove/memorize them, but rather understand why they are true, why it couldn't be any other formula. $\endgroup$ – FuzzyPixelz Jan 10 '17 at 18:32
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    $\begingroup$ @Mahmoud: a proof give you understanding. $\endgroup$ – Arnaldo Jan 10 '17 at 18:33
  • $\begingroup$ @Mahmoud I usually remember it as something derived from the properties of exponential functions. $\endgroup$ – Simply Beautiful Art Jan 10 '17 at 18:38

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