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:

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

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