Funny identities Here is a funny exercise 
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?
 A: \begin{align}\frac{64}{16}&=\frac{6\!\!/\,4}{16\!\!/}\\&=\frac41\\&=4\end{align}
For more examples of these weird fractions, see "How Weird Are Weird Fractions?",
Ryan Stuffelbeam, The College Mathematics Journal, Vol. 44, No. 3 (May 2013), pp. 202-209.
A: Machin's Formula:
\begin{eqnarray}
\frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}.
\end{eqnarray}
A: \begin{eqnarray}
\sum_{k = 0}^{\lfloor q  - q/p) \rfloor} \left \lfloor \frac{p(q - k)}{q} \right \rfloor = \sum_{k = 1}^{q} \left \lfloor \frac{kp}{q} \right \rfloor
\end{eqnarray}
A: $$2592=2^59^2$$ Found this in one of Dudeney's puzzle books
A: $$\frac{\pi}{4}=\sum_{n=1}^{\infty}\arctan\frac{1}{f_{2n+1}}, $$
where $f_{2n+1}$ there are fibonacci numbers, $n=1,2,...$
A: $$\frac{1}{\sin(2\pi/7)} + \frac{1}{\sin(3\pi/7)} = \frac{1}{\sin(\pi/7)}$$
A: The Frobenius automorphism
$$(x + y)^p = x^p + y^p$$
A: \begin{eqnarray}
1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2}
\end{eqnarray}
More generally, let $D_{k} = \{ d\}$ be the set of unitary divisors of a positive integer $k$, and let $\mathsf{d}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the number-of-unitary-divisors (arithmetic) function. Then
\begin{eqnarray}
\sum_{d \in D} \mathsf{d}^{*}(d)^{3} = \left( \sum_{d \in D} \mathsf{d}^{*}(d) \right)^{2}
\end{eqnarray}
Note that $\mathsf{d}^{*}(k) = 2^{\omega(k)}$, where $\omega(k)$ is the number distinct prime divisors of $k$.
A: $$ \sin \theta \cdot \sin \bigl(60^\circ - \theta \bigr) \cdot \sin \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \sin 3\theta$$
$$ \cos \theta \cdot \cos \bigl(60^\circ  - \theta \bigr) \cdot \cos \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \cos 3\theta$$
$$ \tan \theta \cdot \tan \bigl(60^\circ  - \theta \bigr) \cdot \tan \bigl(60^\circ + \theta \bigr) = \tan 3\theta $$
A: $$27\cdot56=2\cdot756,$$
$$277\cdot756=27\cdot7756,$$
$$2777\cdot7756=277\cdot77756,$$
and so on.
A: $$
\sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=1-\frac12-\frac13-\frac15+\frac16-\frac17+\frac1{10}-\frac1{11}-\frac1{13}+\frac1{14}+\frac1{15}-\cdots=0
$$
This relation was discovered by Euler in 1748 (before Riemann's studies on the $\zeta$ function as a complex variable function, from which this relation becomes much more easier!).
Then one of the most impressive formulas is the functional equation for the $\zeta$ function, in its asimmetric form: it highlights a very very deep and smart connection between the $\Gamma$ and the $\zeta$:
$$
\pi^{\frac s2}\Gamma\left(\frac s2\right)\zeta(s)=
\pi^{\frac{1-s}2}\Gamma\left(\frac{1-s}2\right)\zeta(1-s)\;\;\;\forall s\in\mathbb C\;.
$$
Moreover no one seems to have wrote the Basel problem (Euler, 1735):
$$
\sum_{n=1}^{+\infty}\frac1{n^2}=\frac{\pi^2}{6}\;\;.
$$
A: $$\large{1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7}$$
and
$$\large{111,111,111 \times 111,111,111 = 12,345,678,987,654,321}$$
A: $$\sec^2(x)+\csc^2(x)=\sec^2(x)\csc^2(x)$$
A: \[\sqrt{n^{\log n}}=n^{\log \sqrt{n}}\]
A: $\displaystyle\big(a^2+b^2\big)\cdot\big(c^2+d^2\big)=\big(ac \mp bd\big)^2+\big(ad \pm bc\big)^2$
A: $\lnot$(A$\land$B)=($\lnot$A$\lor$$\lnot$B) and
$\lnot$(A$\lor$B)=($\lnot$A$\land$$\lnot$B), because they mean that negation is an "equal form".
A: Here's one clever trigonometric identity that impressed me in high-school days. Add $\sin \alpha$, to both the numerator and the denominator of $\sqrt{\frac{1-\cos \alpha}{1 + \cos \alpha}}$ and get rid of the square root and nothing changes. In other words:
$$\frac{1 - \cos \alpha + \sin \alpha}{1 + \cos \alpha + \sin \alpha} = \sqrt{\frac{1-\cos \alpha}{1 + \cos \alpha}}$$
If you take a closer look you'll notice that the RHS is the formula for tangent of a half-angle. Actually if you want to prove those, nothing but the addition formulas  are required.
A: Facts about $\pi$ are always fun!
\begin{equation}
\frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\ldots\\
\end{equation}
\begin{equation}
\frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\ldots\\
\end{equation}
\begin{equation}
\frac{\pi^2}{6} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\ldots\\
\end{equation}
\begin{equation}
\frac{\pi^3}{32} = 1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\frac{1}{9^3}-\ldots\\
\end{equation}
\begin{equation}
\frac{\pi^4}{90} = 1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots\\
\end{equation}
\begin{equation}
\frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\ldots\\ 
\end{equation}
\begin{equation}
\pi = \cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\ldots}}}}}\\
\end{equation}
A: Well, i don't know whether to classify this as funny or surprising,  but ok it's worth posting.


*

*Let $(X,\tau)$ be a topological space and let $A \subset X$ . By iteratively applying operations of closure and complemention, one can produce at most 14 distinct sets. It's called as the Kuratowski's Closure complement problem. 

A: \begin{align}
\frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1} &\text{Power Rule?}&\ \text{False}\\
\frac{\mathrm d}{\mathrm dx}(x^x) &= x^{x}\ln(x) &\text{Exponential Rule?}&\ \text{False}\\
\frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1}+x^{x}\ln(x) &\text{Sum of these?}&\ \text{True}\\
\end{align}
A: The following number is prime

$p = 785963102379428822376694789446897396207498568951$

and $p$ in base 16 is

$89ABCDEF012345672718281831415926141424F7$

which includes counting in hexadecimal, and digits of $e$, $\pi$, and $\sqrt{2}$.
Do you think this's surprising or not?
$$11 \times 11 = 121$$
$$111 \times 111 = 12321$$
$$1111 \times 1111 = 1234321$$
$$11111 \times 11111 = 123454321$$
$$\vdots$$
A: $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3) = \pi$
(using the principal value),
but if you blindly use the addition formula
$\tan^{-1}(x) + \tan^{-1}(y)
= \tan^{-1}\dfrac{x+y}{1-x y}$
twice, you get zero:
$\tan^{-1}(1) + \tan^{-1}(2) =
\tan^{-1}\dfrac{1+2}{1-1*2}
=\tan^{-1}(-3)$;
$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)
=\tan^{-1}(-3) + \tan^{-1}(3)
=\tan^{-1}\dfrac{-3+3}{1-(-3)(3)}
= 0$.
A: $$
\frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{1/4}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{1/8}\left(\frac{8\cdot 10\cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16}{9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15}\right)^{1/16}\cdots
$$
[Nick Pippenger, Amer. Math. Monthly, 87 (1980)]. Set all the exponents to 1 and you get the Wallis formula for $\pi/2$.
A: $$
\begin{array}{rcrcl}
\vdots & \vdots & \vdots & \vdots & \vdots
\\[1mm]
\int{1 \over x^{3}}\,{\rm d}x & = & -\,{1 \over 2}\,{1 \over x^{2}} & \sim & x^{\color{#ff0000}{\large\bf -2}}
\\[1mm]
\int{1 \over x^{2}}\,{\rm d}x & = & -\,{1 \over x} & \sim & x^{\color{#ff0000}{\large\bf -1}}
\\[1mm]
\int{1 \over x}\,{\rm d}x & = & \ln\left(x\right) & \sim & x^{\color{#0000ff}{\LARGE\bf 0}} \color{#0000ff}{\LARGE\quad ?}
\\[1mm]
\int x^{0}\,{\rm d}x & = & x^{1} & \sim & x^{\color{#ff0000}{\large\bf 1}}
\\[1mm]
\int x\,{\rm d}x & = & {1 \over 2}\,x^{2} & \sim & x^{\color{#ff0000}{\large\bf 2}}
\\[1mm]
\vdots & \vdots & \vdots & \vdots & \vdots
\end{array}
$$
A: $$\lim_{\omega\to\infty}3=8$$ The "proof" is by rotation through $\pi/2$. More of a joke than an identity, I suppose. 
A: For all $n\in\mathbb{N}$ and $n\neq1$
$$\prod_{k=1}^{n-1}2\sin\frac{k \pi}{n} = n$$
For some reason, the proof involves complex numbers and polynomials.
Link to proof: Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$
A: \begin{eqnarray}
\sum_{i_1 = 0}^{n-k} \, \sum_{i_2 = 0}^{n-k-i_1} \cdots \sum_{i_k = 0}^{n-k-i_1 - \cdots - i_{k-1}} 1 = \binom{n}{k}
\end{eqnarray}
A: \begin{align}
E &=
\sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}}
=
mc^{2}
+
\left[\sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} - mc^{2}\right]
\\[3mm]&=
mc^{2}
+
{\left(pc\right)^{2}
 \over
 \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} + mc^{2}}
=
mc^{2}
+
{p^{2}/2m
 \over
 1 + {\sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} - mc^{2} \over 2mc^{2}}}
\\[3mm]&=
mc^{2}
+
{p^{2}/2m
 \over
 1 + {p^{2}/2m \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} + mc^{2}}}
=
mc^{2}
+
{p^{2}/2m
 \over 1 +
{p^{2}/2m \over
1 + {p^{2}/2m \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} - mc^{2}}}}
\end{align}
A: $\textbf{Claim:}\quad$$$\frac{\sin x}{n}=6$$ for all $n,x$ ($n\neq 0$).
$\textit{Proof:}\quad$$$\frac{\sin x}{n}=\frac{\dfrac{1}{n}\cdot\sin x}{\dfrac{1}{n}\cdot n}=\frac{\operatorname{si}x}{1}=\text{six}.\quad\blacksquare$$
A: M.V Subbarao's identity: an integer $n>22$ is a prime number iff it satisfies,
$$n\sigma(n)\equiv 2 \pmod {\phi(n)}$$
A: $${\Large%
\sqrt{\,\vphantom{\huge A}\color{#00f}{20}\color{#c00000}{25}\,}\, =\ \color{#00f}{20}\ +\ \color{#c00000}{25}\ =\ 45}
$$
A: $$\sum\limits_{n=1}^{\infty} n = 1 + 2 + 3 + \cdots \text{ad inf.} = -\frac{1}{12}$$
You can also see many more here: The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation
A: Two related integrals:
$$\int_0^\infty\sin\;x\quad\mathrm{d}x=1$$
$$\int_0^\infty\ln\;x\;\sin\;x\quad \mathrm{d}x=-\gamma$$
A: $$ 10^2+11^2+12^2=13^2+14^2 $$
There's a funny Abstruse Goose comic about this, which I can't seem to find at the moment.
A: $32768=(3-2+7)^6 / 8$
Just a funny coincidence.
A: By excluding the first two primes, Euler's Prime Product becomes a square:
$$\prod _{n=3}^{\infty } \frac{1}{1-\frac{1}{(p_n)^{2}}}=\frac{\pi ^2}{9}$$
By using multiples of the product of the first two primes, we get the square root:
$$\prod _{n=1}^{\infty } \frac{1}{1-\frac{1}{(n p_1 p_2)^{2}}}=\frac{\pi }{3}$$
A: I have one: In a $\Delta ABC$,
$$\tan A+\tan B+\tan C=\tan A\tan B\tan C.$$
A: $$\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty \frac1{k^k}$$
A: 
$$\left|z+z'\right|^{2}+\left|z-z'\right|^{2}=2\times\left(\left|z\right|^{2}+\left|z'\right|^{2}\right)$$

The sum of the squares of the sides
  equals the sum of the squares of the
  diagonals.

A: What is 42?
$$
6 \times 9 = 42 \text{ base } 13
$$
I always knew that there is something wrong with this universe.
A: The product of any four consecutive integers is one less than a perfect square.
To phrase it more like an identity:
For every integer $n$, there exists an integer $k$ such that
$$n(n+1)(n+2)(n+3) = k^2 - 1.$$
A: Considering the main branches
$$i^i = \exp\left(-\frac{\pi}{2}\right)$$
$$\root i \of i  = \exp\left(\frac{\pi}{2}\right) $$
And 
$$ \frac{4}{\pi } = \displaystyle 1 + \frac{1}{{3 +\displaystyle \frac{{{2^2}}}{{5 +  \displaystyle\frac{{{3^2}}}{{7 +\displaystyle \frac{{{4^2}}}{{9 +\displaystyle \frac{{{n^2}}}{{\left( {2n + 1} \right) +  \cdots }}}}}}}}}} $$
A: $$
\frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{2^{2k}}
$$
$$
\frac{\pi}{3}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{6^{2k}}
$$
where 
$
\eta(n)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{n}}
$
A: $$\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^\pi}dx=\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^e}dx$$
A: Let $f$ be a symbol with the property that $f^n = n!$. Consider $d_n$, the number of ways of putting $n$ letters in $n$ envelopes so that no letter gets to the right person (aka derangements). Many people initially think that $d_n = (n-1)! = f^{n-1}$ (the first object has $n-1$ legal locations, the second $n-2$, ...). The correct answer isn't that different actually:
$d_n = (f-1)^n$. 
A: Best near miss
$$\int_{0}^{\infty }\cos\left ( 2x \right )\prod_{n=0}^{\infty}\cos\left ( \frac{x}{n} \right )~\mathrm dx\approx \frac{\pi}{8}-7.41\times 10^{-43}$$
One can easily be fooled into thinking that it is exactly $\dfrac{\pi}{8}$.
References:


*

*Wikipedia

*Future Prospects for Computer-Assisted Mathematics, by D.H. Bailey and J.M. Borwein

A: $$\left(\sum\limits_{k=1}^n k\right)^2=\sum\limits_{k=1}^nk^3 .$$
The two on the left is not a typo.
A: I actually think currying is really cool:
$$(A \times B) \to C \; \simeq \; A \to (B \to C)$$
Though not strictly an identity, but an isomorphism.
When I met it for the first time it seemed to be a bit odd but it is so convenient and neat. At least in programming.
A: $$ 71 = \sqrt{7! + 1}. $$
Besides the amusement of reusing the decimal digits $7$ and $1$,
this is conjectured to be the last solution of $n!+1 = x^2$ in integers.
($n=4$ and $n=5$ also work.)  Even finiteness of the set of solutions
is not known except using the ABC conjecture.
A: We have by block partition rule for determinant 
$$
\det 
\left[
\begin{array}{cc}
U & R \\ 
L & D 
\end{array}
\right]
=
\det U\cdot \det ( D-LU^{-1}R)
$$
But if $U,R,L$ and $D$ commute we have that
$$
\det 
\left[
\begin{array}{cc}
U & R \\ 
L & D 
\end{array}
\right]
=
\det (UD-LR)
$$
A: Heres a interesting one again
$3435=3^3+4^4+3^3+5^5%$
A: The Cayley-Hamilton theorem:
If $A \in \mathbb{R}^{n \times n}$ and $I_{n} \in \mathbb{R}^{n \times n}$ is the identity matrix, then the characteristic polynomial of $A$ is $p(\lambda) = \det(\lambda I_n - A)$. Then the Cayley Hamilton theorem can be obtained by "substituting" $\lambda = A$, since $$p(A) = \det(AI_n-A) = \det(0-0) = 0$$
A: $(x-a)(x-b)(x-c)\ldots(x-z) = 0$
A: $$\frac{1}{998901}=0.000001002003004005006...997999000001...$$
A: $$ \infty! = \sqrt{2 \pi} $$
It comes from the zeta function.
A: \begin{eqnarray}
\zeta(0) = \sum_{n \geq 1} 1 = -\frac{1}{2}
\end{eqnarray}
A: $$
\dfrac{1}{2}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\cdots}}}}}}
$$
and more generally we have
$$
\dfrac{1}{n+1}=\frac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\frac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\ddots}}}}}}
$$
A: Ah, this is one identity which comes into use for proving the Euler's Partition Theorem. The identity is as follows: $$ (1+x)(1+x^{2})(1+x^{3}) \cdots  = \frac{1}{(1-x)(1-x^{3})(1-x^{5}) \cdots}$$
A: If we define $P$ as the infinite lower triangular matrix where $P_{i,j} = \binom{i}{j}$ (we can call it the Pascal Matrix), then $$P^k_{i,j} = \binom{i}{j}k^{i-j}$$
where $P^k_{i,j}$ is the element of $P^k$ in the position $i,j.$
A: Voronoi summation formula:
$\sum \limits_{n=1}^{\infty}d(n)(\frac{x}{n})^{1/2}\{Y_1(4\pi \sqrt{nx})+\frac{2}{\pi}K_1(4\pi \sqrt{nx})\}+x \log x +(2 \gamma-1)x +\frac{1}{4}=\sum \limits _{n\leq x}'d(n)$
A: Let $\sigma(n)$ denote the sum of the divisors of $n$.
If $$p=1+\sigma(k),$$ then $$p^a=1+\sigma(kp^{a-1})$$
where $a,k$ are positive integers and $p$ is a prime such that $p\not\mid k$.
A: I have another one, but I'm quite unwilling to post this here because it's MINE, I haven't found it anywhere, so don't steal this.
Let us take the four most important mathematical constants: The Euler number $e$, the Aurea Golden Ratio $\phi$, the Euler-Mascheroni constant $\gamma$ and finally $\pi$. Well we can see easily that
$$e\cdot\gamma\cdot\pi\cdot\phi \approx e + \gamma + \pi + \phi$$
A: $$
\int_{-\infty}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x
=
\pi\int_{-1}^{1}\delta\left(k\right)\,{\rm d}k
$$
