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### Euler's identity for $\frac{sin(x)}{x}$

Here is this famous identity due to Euler.

$$\frac{\sin x}{x} = \prod_{k=1}^{\infty} \cos \left(\frac{x}{2^k} \right) \tag{1}$$

This holds true for every $x \ne 0$.

Let's prove it.

Denote:

$$S(n) = \prod_{k=1}^{n} \cos \left(\frac{x}{2^k} \right) \tag{2}$$

Multiplying the two sides by $sin{\frac{x}{2^n}}$ we sequentially get:

$$sin{\frac{x}{2^n}} \cdot S(n) = sin{\frac{x}{2^{n}}} \cdot \prod_{k=1}^{n} \cos \left(\frac{x}{2^k} \right)$$
$$sin{\frac{x}{2^n}} \cdot S(n) = sin{\frac{x}{2^{n-1}}} \cdot \frac{1}{2^1} \cdot \prod_{k=1}^{n-1} \cos \left(\frac{x}{2^k} \right)$$
$$sin{\frac{x}{2^n}} \cdot S(n) = sin{\frac{x}{2^{n-2}}} \cdot \frac{1}{2^2} \cdot \prod_{k=1}^{n-2} \cos \left(\frac{x}{2^k} \right)$$
$$...$$
$$sin{\frac{x}{2^n}} \cdot S(n) = sin{\frac{x}{2^{1}}} \cdot \frac{1}{2^{n-1}} \cdot \prod_{k=1}^{1} \cos \left(\frac{x}{2^k} \right)$$

The last one obviously gives us:

$$sin{\frac{x}{2^n}} \cdot S(n) = \frac{1}{2^{n}} \cdot sin(x) \tag{2}$$

which can be easily reworked to:

$$\large{ \frac{sin{\frac{x}{2^n}}}{\frac{x}{2^n}} \cdot S(n) = \frac{sin(x)}{x}} \tag{3}$$

Now in $(3)$ when we take the limit as ${n\to\infty}$ while using that

$$\lim_{u \to 0} \frac{sin(u)}{u} = 1$$

we get equality $(1)$ which is what we wanted to prove.

### Trigonometric identities

1) Even/Odd function identities

$$\sin(-\alpha) = -\sin\alpha \tag{1.1}$$
$$\cos(-\alpha) = \cos\alpha \tag{1.2}$$
$$\tan(-\alpha) = -\tan\alpha \tag{1.3}$$
$$\cot(-\alpha) = -\cot\alpha \tag{1.4}$$

$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \tag{2.1}$$
$$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta \tag{2.2}$$
$$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \tag{2.3}$$
$$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta \tag{2.4}$$

$$\tan(\alpha + \beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\tag{2.5}$$

$$\cot(\alpha + \beta) = \frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}\tag{2.6}$$

3) Sum to product identities

$$\sin\alpha + \sin\beta = 2 \sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \tag{3.1}$$
$$\sin\alpha - \sin\beta = 2 \sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2} \tag{3.2}$$

$$\cos\alpha + \cos\beta = 2 \cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \tag{3.3}$$

$$\cos\alpha - \cos\beta = -2 \sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} \tag{3.4}$$

$$\tan\alpha \pm \tan\beta = \frac{\sin{(\alpha\pm\beta)}}{\cos\alpha\cos\beta} \tag{3.5}$$
$$\cot\alpha \pm \cot\beta = \frac{\sin{(\beta\pm\alpha)}}{\sin\alpha\sin\beta} \tag{3.6}$$

4) Product to sum identities

$$\sin\alpha\sin\beta = \frac{1}{2}[\cos(\alpha-\beta) - \cos(\alpha+\beta)] \tag{4.1}$$

$$\cos\alpha\cos\beta = \frac{1}{2}[\cos(\alpha-\beta) + \cos(\alpha+\beta)] \tag{4.2}$$

$$\sin\alpha\cos\beta = \frac{1}{2}[\sin(\alpha+\beta) + \sin(\alpha-\beta)] \tag{4.3}$$

5) Double-angle and triple-angle identities

$$\sin 2 \alpha = 2 \sin\alpha \cos\alpha \tag{5.1}$$

$$\cos 2 \alpha = \cos^2\alpha - \sin^2\alpha \tag{5.2}$$

$$\tan 2 \alpha = \frac{2\tan\alpha}{1-\tan^2\alpha} = \frac{2}{\cot\alpha - \tan\alpha} \tag{5.3}$$

$$\cot 2 \alpha = \frac{\cot^2\alpha-1}{2\cot\alpha} = \frac{\cot\alpha - \tan\alpha}{2} \tag{5.4}$$

$$\sin 3 \alpha = 3 \sin\alpha - 4\sin^3 \alpha \tag{5.5}$$

$$\cos 3 \alpha = 4\cos^3 \alpha - 3 \cos\alpha \tag{5.6}$$

$$\tan 3 \alpha = \frac{3\tan \alpha - \tan^3 \alpha}{1 - 3 \tan^2 \alpha} \tag{5.7}$$

$$\cot 3 \alpha = \frac{\cot^3 \alpha - 3 \cot \alpha}{3\cot^2 \alpha - 1} \tag{5.8}$$

6) Decreasing the power of sine and cosine

$$2 \sin^2 \alpha = 1 - \cos 2\alpha \tag{6.1}$$

$$4 \sin^3 \alpha = 3 \sin \alpha - \sin 3\alpha \tag{6.2}$$

$$8 \sin^4 \alpha = \cos 4\alpha - 4 \cos 2\alpha + 3 \tag{6.3}$$

$$2 \cos^2 \alpha = 1 + \cos 2\alpha \tag{6.4}$$

$$4 \cos^3 \alpha = 3 \cos \alpha + \cos 3\alpha \tag{6.5}$$

$$8 \cos^4 \alpha = \cos 4\alpha + 4 \cos 2\alpha + 3 \tag{6.6}$$

### Fun problem on Ceva's theorem

Problem:

Given is $\Delta ABC$. The points $D \in BC$, $E \in CA$, $F \in AB$ are such that the lines $AD,BE,CF$ are concurrent.

$A'$ - midpoint of $BC$
$B'$ - midpoint of $CA$
$C'$ - midpoint of $AB$

$D'$ - midpoint of $AD$
$E'$ - midpoint of $BE$
$F'$ - midpoint of $CF$

Prove that the lines

$A'D', B'E', C'F'$ are also concurrent (i.e. that they pass through a common point).

Solution:

It is crucial to come up with a nice realistic drawing here.

Also, it is important to realize that:
a) $A', F', B'$ are on one line
b) $B', D', C'$ are on one line
c) $C', E', A'$ are on one line

From Ceva's theorem for triangle $ABC$ we get:
$$\frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA} = 1 \tag{1}$$

Now the trick is to realize that:
$$\frac{B'F'}{F'A'} = \frac{AF}{FB} \tag{2}$$
$$\frac{C'D'}{D'B'} = \frac{BD}{DC} \tag{3}$$
$$\frac{A'E'}{E'C'} = \frac{CE}{EA} \tag{4}$$

Why is this so?

Because $B'C' || BC$   , $C'A' || CA$ and  $A'B' || AB$
so these relations follow from the Intercept theorem.

Multiplying the last 3 equations and using $(1)$ we get:

$$\frac{B'F'}{F'A'}\frac{C'D'}{D'B'}\frac{A'E'}{E'C'} = \frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA} = 1$$

Thus:

$$\frac{B'F'}{F'A'}\frac{A'E'}{E'C'}\frac{C'D'}{D'B'} = 1 \tag{5}$$

Now using the converse Ceva's theorem (for the triangle $A'B'C'$ and for the points $D', E', F'$), we can conclude from $(5)$ that the three lines  $A'D', B'E', C'F'$ intersect at a single/common point. This is what we had to prove hence the problem is solved.

### Klein bottle mystery

In the book "Basic Topology" by M.A.Armstrong I found an explanation about how to construct a Klein bottle. I had to reread it 5 times and I was still not quite convinced. I am retelling it here.

Begin with a sphere, remove two discs from it, and add a Möbius strip in their places.
A Möbius strip has after all a single circle as boundary, and all that we are asking
is that the points of its boundary circle be identified with those of the boundary
circle of the hole in the sphere. One must imagine this identification taking place
in some space where there is plenty of room (euclidean four-dimensional space will do).
This cannot be realized in three dimensions without having each Möbius strip
intersect itself. The resulting closed surface is called the Klein bottle.

I was scratching my head around how this procedure actually produces a Klein bottle until I found this question in MathSE.

Klein-bottle-as-two-Möbius-strips

This picture in one of the answers is really really nice, it really shows what happens if we cut a Klein bottle in half - we really get two Möbius strips as a result. The cut is done by a plane "parallel to the handle" which cuts the bottle in two symmetric parts.

So... it's really for a reason that they say "a picture is worth a thousand words".

### How to solve a quadratic equation?

This post comes to demonstrate the support for $\LaTeX$ available in Blogger.

A quadratic equation is an equation of the form
$$ax^2 + bx + c = 0 \tag{1}$$
where
$a,b,c \in \mathbb{R}$ and $a \ne 0$.
Let us assume that we are trying to solve this equation for real numbers only.
The value
$$D = b^2 - 4ac \tag{2}$$
is called discriminant of the equation $(1)$.
Case #1:
When $D \gt 0$, there are two distinct real solutions to $(1)$ and they are:
$$x_{1,2} = {-b \pm \sqrt{b^2-4ac} \over 2a} \tag{3}$$
Case #2:
When $D = 0$, there is one real solution to $(1)$ which is:
$$x = {-b \over 2a} \tag{4}$$
Case #3:
When $D \lt 0$, there are no real solutions to $(1)$.

### How to solve a linear equation?

This post comes to demonstrate the support for $\LaTeX$ available in Blogger.

A linear equation is an equation of the form

$$ax + b = 0 \tag{1}$$

where

$$a,b \in \mathbb{R}, a \ne 0$$

Let us assume that we are trying to solve this equation for real numbers only i.e. we are trying to find the real roots of the equation.

Solving this equation is simple, it always has a single real solution which is

$$x = - {b \over a} \tag{2}$$

### Calculate $\lim_{x \to \infty} \sqrt[n]{(x+a_1)(x+a_2) ... (x+a_n)} - x$

This is a problem which I found in a math textbook some time ago. I tried solving it but I did not manage to solve it as quickly as I wanted and so I lost patience. Therefore I posted it in MathSE. I liked quite a lot one of the solutions I got, so I wrote it on a sheet of paper and kept it.

Here is the full story...

Problem:

Find $\lim_{x \to \infty} \sqrt[n]{(x+a_1)(x+a_2) ... (x+a_n)} - x$,

where $x$ is a real variable and $n$ is a fixed natural number $n \ge 1$.

Solution:

Let
$f(x)=\sqrt[n]{(x+a_1)(x+a_2)\cdot (x+a_n)}$

We now calculate:

$$\lim_{x\to \infty} \frac{f(x)}{x}=\lim_{x\to \infty} \sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x}) \cdot (1+\frac{a_n}{x})}= \sqrt[n]{1} = 1 \tag{1}$$

We now form the below difference and we rework it a bit:

$$f(x)-x=\frac{f(x)^n-x^n}{\sum_{k=0}^{n-1} f(x)^{n-1-k} x^{k}}= \frac{f(x)^n-x^n}{f(x)^{n-1}+f(x)^{n-2}x+\dots+f(x)x^{n-2}+x^{n-1}}$$

We note here that the numerator $P(x) = f(x)^n-x^n$ is a polynomial of degree $n-1$ and it has a leading coefficient of $c = a_1+a_2+\dots+a_n$. The denominator is composed of the sum of $n$ terms of the form $f(x)^{n-1-k}x^k$ (for $k=0,1, \dots, n-1$).

So we divide the numerator and the denominator by $x^{n-1}$ and we get:

\begin{align*}
\smash
\lim_{x\to \infty}f(x)-x & = \lim_{x\to \infty} \frac {\frac{f(x)^n-x^n}{x^{n-1}}}{\frac{f(x)^{n-1}}{x^{n-1}}+\frac{f(x)^{n-2}}{x^{n-2}}+\dots+\frac{f(x)}{x}+1} \\[10pt]
&= \lim_{x\to \infty} \frac {\frac{P(x)}{x^{n-1}}}{\frac{f(x)^{n-1}}{x^{n-1}}+\frac{f(x)^{n-2}}{x^{n-2}}+\dots+\frac{f(x)}{x}+1} \\[10pt]
&= \frac{c}{\underbrace{1 + 1 + \dots + 1}_{n\text{ times}}} \\[10pt]
&= \frac{a_1+a_2+\dots+a_n}{n}\end{align*}

The cool thing here is that the limit value happens to be the arithmetic mean of the numbers $a_1, a_2, \dots a_n$.

### Derivative of $f(x)=x^n$

What is the derivative of $$f(x)=x^n$$ when $n$ is a fixed integer and $x$ is a real variable?

As we all know this derivative is

$$f'(x)=nx^{n-1}$$

But for which values of $n$ and $x$ does this hold true?

Well...

1) this is true for integers $n \gt 0$
2) this is also true for integers $n \lt 0$ if $x \ne 0$

Note that when $x=0$ the symbols $x^0, x^{-1}, x^{-2}, x^{-3}, \dots$ are usually treated as undefined. That is why for 2) we need the additional restriction that $x \ne 0$.