heureka

the expression cot times sec is equivalent to what

Weierstrass substitution:

$$\\\small{\text{
$
\boxed{
\rm{if~}t = \tan\tfrac{x}{2} \rm{~then~}
}$}}\\
\small{\text{
$
\boxed{
\sin x = \frac{2t}{1 + t^2},~
\cos x = \frac{1 - t^2}{1 + t^2},~
\tan x = \frac{2t}{1 - t^2},~
\cot x = \frac{1 - t^2}{2t},~
\sec x = \frac{1 + t^2}{1 - t^2},~
\csc x = \frac{1 + t^2}{2t}.
}$}}$$

$$\cot{(x)} \cdot \sec {(x)} =
\left( \frac{1 - t^2}{2t} \right)
\cdot
\left( \frac{1 + t^2}{1 - t^2} \right)
= \frac{1 + t^2}{2t} = \csc{( x )}$$

$$\small{\text{
$
\lim \limits_{n \rightarrow \infty } \sqrt{n}\cdot
\left(
\sqrt{n^2+1}-n
\right)
$}}\\\\
=
\small{\text{
$
\lim \limits_{n \rightarrow \infty } \sqrt{n}\cdot
\left(
\sqrt{\frac{n^2}{n^2}\cdot(n^2+1)}-n
\right)
$}}\\\\
=
\small{\text{
$
\lim \limits_{n \rightarrow \infty } \sqrt{n}\cdot
\left(n\cdot
\sqrt{ 1+\frac{1}{n^2} }-n
\right)
$}}\\\\
=
\small{\text{
$
\lim \limits_{n \rightarrow \infty } \sqrt{n}\cdot n\cdot
\left(
\sqrt{ 1+\frac{1}{n^2} }-1
\right)
$}}\\\\
=
\small{\text{
$
\lim \limits_{n \rightarrow \infty } \sqrt{n}\cdot n\cdot
\left(
\sqrt{ 1+\frac{1}{n^2} } - 1
\right) \dfrac{ \sqrt{ 1+\frac{1}{n^2} } + 1 }
{ \sqrt{ 1+\frac{1}{n^2} } + 1 }
$}}\\\\
=
\small{\text{
$
\lim \limits_{n \rightarrow \infty } \sqrt{n}\cdot n\cdot
\left(
1+\frac{1}{n^2} - 1^2
\right) \dfrac{ 1 }
{ \sqrt{ 1+\frac{1}{n^2} } + 1 }
$}}\\\\
=
\small{\text{
$
\lim \limits_{n \rightarrow \infty }
\dfrac{ \sqrt{n}\cdot n } { n^2 } \cdot
\dfrac{ 1 } { \sqrt{ 1+\frac{1}{n^2} } + 1 }
$}}\\\\
=
\small{\text{
$
\lim \limits_{n \rightarrow \infty }
\dfrac{ \dfrac{ \sqrt{n} } { n } }
{ \sqrt{ 1+\frac{1}{n^2} } + 1 }
$}}\\\\
=
\small{\text{
$
\lim \limits_{n \rightarrow \infty }
\dfrac{ \sqrt{ \dfrac{ n } { n^2 } } }
{ \sqrt{ 1+\frac{1}{n^2} } + 1 }
$}}$$

$$\\=
\small{\text{
$
\lim \limits_{n \rightarrow \infty }
\dfrac{ \sqrt{ \dfrac{ 1} { n } } }
{ \sqrt{ 1+\frac{1}{n^2} } + 1 }
$}}\\\\
=
\small{\text{
$
\lim \limits_{n \rightarrow \infty }
\dfrac{ \sqrt{ \dfrac{ 1} { n } } }
{ \sqrt{ 1+ (\frac{1}{n})^2 } + 1 }
$}} \qquad | \qquad
\boxed{~
\lim \limits_{n \rightarrow \infty } \dfrac{1}{n}=0~}\\\\
=
\small{\text{
$ \dfrac{ \sqrt{0} }
{ \sqrt{ 1 + 0^2 } + 1 }
$}}\\\\
=
\small{\text{
$ \dfrac{ 0 }
{ \sqrt{ 1 } + 1 }
$}}\\\\
=
\small{\text{
$ \dfrac{ 0 }
{ 1 + 1 }
$}}\\\\
=
\small{\text{
$ \dfrac{ 0 }
{ 2 }
$}}\\\\
=
\small{\text{$ 0 $}}$$

2620mm sind wie viel meter ?

$$\small{\text{
$
2620~\rm{mm}
\cdot \dfrac{ 1~\rm{cm} }{ 10~\rm{mm} }
\cdot \dfrac{ 1~\rm{dm} }{ 10~\rm{cm} }
\cdot \dfrac{ 1~\rm{m} }{ 10~\rm{dm} }
$}}\\\\\\
\small{\text{
$
\begin{array}{rcl}
&=& 2620
\cdot \dfrac{1}{10}
\cdot \dfrac{1}{10}
\cdot \dfrac{1}{10}
\cdot \dfrac{ \rm{mm} }{ \rm{mm} }
\cdot \dfrac{ \rm{cm} }{ \rm{cm} }
\cdot \dfrac{ \rm{dm} }{ \rm{dm} }
\cdot \rm{m} \qquad |
\qquad \dfrac{ \rm{mm} }{ \rm{mm} }=1
\qquad \dfrac{ \rm{cm} }{ \rm{cm} }=1
\qquad \dfrac{ \rm{dm} }{ \rm{dm} }=1 \\\\\\
&=& \dfrac{2620}{ 10 \cdot 10\cdot 10 } \cdot \rm{m}\\\\\\
&=& \dfrac{2620}{ 1000 } \cdot \rm{m} \\\\\\
&=& 2,620 \cdot \rm{m}
\end{array}
$}}$$

how to represent -77,626 in binary form ?

$$\begin{array}{rcrr}
& & \rm{q o u t i e n t} & \rm{r e m a i n d e r} \\
77626& : 2 = & 38813 & 0 \\
38813 & : 2 = & 19406 & 1 \\
19406 & : 2 = & 9703& 0 \\
9703 & : 2 = & 4851& 1 \\
4851 & : 2 = & 2425& 1 \\
2425 & : 2 = & 1212 & 1 \\
1212 & : 2 = & 606 & 0 \\
606 & : 2 = & 303 & 0 \\
303 & : 2 = & 151 & 1 \\
151 & : 2 = & 75 & 1 \\
75 & : 2 = & 37 & 1 \\
37 & : 2 = & 18 & 1 \\
18 & : 2 = & 9 & 0 \\
9 & : 2 = & 4 & 1 \\
4 & : 2 = & 2 & 0 \\
2 & : 2 = & 1 & 0 \\
1 & : 2 = & 0 & \textcolor[rgb]{1,0,0}{1} \\
\end{array}$$

$$\begin{array}{rcl}
77626_2~&=& \textcolor[rgb]{0,0,1}{0}~\textcolor[rgb]{1,0,0}{1}~0~0~1~0~1~1~1~1~0~0~1~1~1~0~1~0\\
\mathbf{Invert~all~places} &=& \textcolor[rgb]{0,0,1}{1}~\textcolor[rgb]{1,0,0}{0}~1~1~0~1~0~0~0~0~1~1~0~0~0~1~0~1\\
\mathbf{+1} &=+ & \textcolor[rgb]{0,0,1}{0}~0~0~0~0~0~0~0~0~0~0~0~0~0~0~0~0~1\\
-77626_2~&=& \textcolor[rgb]{0,0,1}{1}~0~1~1~0~1~0~0~0~0~1~1~0~0~ 0~1~1~0
\end{array} \\\\
\textcolor[rgb]{0,0,1}{\mathrm{sign: \qquad 0 = + \qquad 1 = -}}$$

Express  as a common fraction.

$$\left\begin{array}{rcl}
10 \times 3.\overline{7} &=& 37.\overline{7} \\
1 \times 3.\overline{7} &=& 3.\overline{7} \\
\hline
\end{array} \right\} -$$

$$\begin{array}{rcl}
(10-1)\times 3.\overline{7} &=& 37.\overline{7} - 3.\overline{7} \\
9 \times 3.\overline{7} &=& 34 \\
9 \times 3.\overline{7} &=& 34 \qquad | \qquad : 9 \\
3.\overline{7} &=& \dfrac{34}{9}
\end{array}$$

25*2^x-8*5^(x-1)=0

$$\small{\text{
$
\begin{array}{rcl}
25\cdot 2^x -8\cdot 5^{x-1} &=& 0 \\
25\cdot 2^x &=& 8\cdot 5^{x-1} \\
5^2\cdot 2^x &=& 2^3\cdot 5^{x-1} \\\\
\dfrac{2^x}{2^3} &=&\dfrac{5^{x-1}}{5^2} \\\\
2^{x-3} &=& 5^{x-1-2} \\
2^{x-3} &=& 5^{x-3} \\\\
\dfrac{ 2^{x-3} } {5^{x-3}} &=& 1\\\\
\left( \dfrac{ 2 }{ 5 } \right)^{x-3} &=& 1\\\\
0,4^{x-3} &=& 1\\
0,4^{x-3} &=& 1 \qquad | \qquad 1 = 0,4^0\\
0,4^{x-3} &=& 0,4^0 \qquad | \qquad \mathbf{Koeffizientenvergleich}\\
x-3 &=& 0 \\
x &=& 3
\end{array}
$}}$$

Find 

$$\left
\begin{array}{rcl}
10 \times 0.\overline{9} &=& 9.\overline{9} \\
1 \times 0.\overline{9} &=& 0.\overline{9}
\end{array} \right\} -
\\
\hline$$

$$\begin{array}{rcl}
(10-1)\times 0.\overline{9} &=& 9.\overline{9} - 0.\overline{9} \\
9 \times 0.\overline{9} &=& 9 \\
9 \times 0.\overline{9} &=& 9 \qquad | \qquad : 9 \\
0.\overline{9} &=& \dfrac{9}{9} \\
0.\overline{9} &=& 1 \\
\end{array}$$

 = 1 - 1 = 0

how many decimeters is a meter .  1 m = 10 dm

Bei einer zweistelligen Zahl ist der Zehner doppelt so groß wie der Einer. Vertauscht man Zehner und Einer, so wird die Zahl um 27 kleiner.

$$\small{\text{$
\begin{array}{lrcl}
(1)\qquad & x\cdot 10 + y &=& z \\
(2)\qquad & 2\cdot y &=& x \qquad | \qquad y = \dfrac{x}{2}\\
\hline
\\
(2) \rm{~in~} ( 1) &x\cdot 10 + \dfrac{x}{2} &=& z \\ \\
& \dfrac{21}{2}\cdot x &=& z \\\\
\hline
\\
(3) \qquad & y \cdot 10 + x &=& z - 27 \\\\
(2) \rm{~in~} ( 3) & \dfrac{x}{2} \cdot 10 + x &=& \dfrac{21}{2}\cdot x - 27 \\ \\
& 5 \cdot x + x &=& \dfrac{21}{2}\cdot x - 27\\\\
& 6 \cdot x &=& \dfrac{21}{2}\cdot x - 27 \qquad | \qquad \cdot 2 \\\\
& 12 \cdot x &=& 21 \cdot x - 54 \\\\
& 21 \cdot x - 12 \cdot x &=& 54 \\\\
& 9 \cdot x &=& 54 \\\\
& x &=& 6 \\\\
\hline
\\
& y &=& \dfrac{x}{2} \\\\
& y &=& \dfrac{6}{2} \\ \\
& y &=& 3
\end{array}
$}}$$

$$\begin{array}{rcl}
63 \\
\Downarrow\\
36 \\
\hline
63-27 = 36
\end{array}$$

Both circles have a radius  r=1.  Can you make these 3 regions,  A  B  and  C, to occupy the equal areas ?

$$\small{\text{$
\rm{central~angle}= \varphi_\rm{rad}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\rm{area}_B &=&
2\cdot\left[
\pi r^2 \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_{\rm{rad}} )}
\right] \\
\rm{area}_A &=& \rm{area}_B=\rm{area}_C\\\\
\rm{area}_{circle_1} = \rm{area}_{circle_2} &=& \pi r^2\\
\rm{area}_{circle_1} - \rm{area}_B &=& \rm{area}_B \\
\rm{area}_{circle_1} &=& 2\cdot \rm{area}_B \\
\pi r^2 &=& 2\cdot \rm{area}_B \\
\pi r^2 &=& 2\cdot 2\cdot\left[
\pi r^2 \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_{\rm{rad}} )} \right] \\
\pi r^2 &=& 2\cdot r^2 \cdot \varphi_{\rm{rad}}
- 2 \cdot r^2 \cdot
\sin{ ( \varphi_{\rm{rad}} ) } \qquad | \qquad : r^2\\
\pi &=& 2 \cdot \varphi_{\rm{rad}} - 2 \cdot \sin{ ( \varphi_{\rm{rad}} ) } \qquad | \qquad : 2\\
\frac{\pi}{2} &=& \varphi_{\rm{rad}} - \sin{ ( \varphi_{\rm{rad}} ) } \\\\
\end{array}
$}}$$

$$\boxed{
\varphi_{\rm{rad}} - \sin{ ( \varphi_{\rm{rad}} ) } = \frac{\pi}{2}
}$$

The Newton-Raphson method in one variable is implemented as follows:

Given a function ƒ defined over the reals x, and its derivative ƒ',

we begin with a first guess x0 for a root of the function f.

Provided the function satisfies all the assumptions made in the derivation of the formula,

a better approximation x1 is

$$\small{\text{
$
x_{1} = x_0 - \frac{f(x_0)}{f'(x_0)} \,$.
}}$$

$$\varphi_{\rm{rad}_1} =
\varphi_{\rm{rad}_0} - \dfrac{ \varphi_{\rm{rad}_0} -\sin{ ( \varphi_{\rm{rad}_0} ) } - \frac{\pi}{2}
}
{1-\cos{ (\varphi_{\rm{rad}_0}) } }$$

Iteration:

$$\small{\text{$
\begin{array}{rcl}
\varphi_{\rm{rad}_0} &=& 2\\
\varphi_{\rm{rad}_1} &=& 2.33901410590 \\
\varphi_{\rm{rad}_2} &=& 2.31006319657 \\
\varphi_{\rm{rad}_3} &=& 2.30988146730 \\
\varphi_{\rm{rad}_4} &=& 2.30988146001 \\
\cdots \\
\varphi_{\rm{rad}} = 2.30988146001 \\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\rm{area}_B &=&
2\cdot\left[
\pi r^2 \cdot \frac{\varphi_\rm{rad} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_\rm{rad} )}
\right] \qquad | \qquad r=1 \\
\rm{area}_B &=&
2\cdot\left[
\pi \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot \sin{(\varphi_{\rm{rad}} )}
\right] \\
\rm{area}_B &=& \varphi_{\rm{rad}} - \sin{(\varphi_{\rm{rad}} )} \\\\
\rm{area}_B = 2.30988146001 -\sin{ ( 2.30988146001 )} &=& \frac{\pi}{2}\\
\rm{area}_A = \rm{area}_C &=& \pi r^2 - \rm{area}_B
\qquad | \qquad r=1 \\
\rm{area}_A = \rm{area}_C &=& \pi - \frac{\pi}{2} = \frac{\pi}{2}
\end{array}
$}}$$