heureka

cos(-pi/6) csc(-pi/3) tan(-pi/6) = ?

$\begin{array}{rcll} \cos( \frac{-\pi}{6} )\cdot \csc( \frac{-\pi}{3} ) \cdot\tan( \frac{-\pi}{6} ) &&\\\\ =\cos( \frac{-\pi}{6} )\cdot \frac{1}{\sin( \frac{-\pi}{3} )}\cdot \tan( \frac{-\pi}{6} )\\\\ \boxed{~ \cos( \frac{-\pi}{6} ) = +\cos( \frac{\pi}{6} ) = \frac{1}{2}\sqrt{3} \\ \sin( \frac{-\pi}{3} ) = -\sin( \frac{\pi}{3} ) = -\frac{1}{2}\sqrt{3} \\ \tan( \frac{-\pi}{6} ) = -\tan( \frac{\pi}{6} ) = -\frac{1}{3}\sqrt{3} ~}\\\\ =( \frac{1}{2}\sqrt{3} ) \cdot \left( \frac{1}{-\frac{ 1 }{2}\sqrt{3}} \right)\cdot (-\frac{1}{3}\sqrt{3})\\\\ =\left( \frac{ \frac{1}{2}\sqrt{3} }{-\frac{ 1 }{2}\sqrt{3}} \right)\cdot (-\frac{1}{3}\sqrt{3})\\\\ =( -1 )\cdot (-\frac{1}{3}\sqrt{3})\\\\ =\frac{1}{3}\sqrt{3}\\\\ \mathbf{\cos( \frac{-\pi}{6} )\cdot \csc( \frac{-\pi}{3} ) \cdot\tan( \frac{-\pi}{6} ) =\frac{1}{3}\sqrt{3}} \end{array}$

dB = 20*Log(p1/p2) I need to solve for p1 what is the formula?

$\begin{array}{rcll} dB &=& 20\cdot \log{ \left( \frac{p_1}{p_2} \right) }\\\\ dB &=& \log{ [ \left( \frac{p_1}{p_2} \right)^{20} ] } \qquad &| \qquad 10^{()}\\\\ 10^{dB} &=&10^{ \log{ [ \left( \frac{p_1}{p_2} \right)^{20} ] } } \\\\ 10^{dB} &=& \left( \frac{p_1}{p_2} \right)^{20} \qquad &| \qquad 1^{\frac{1}{20}}\\\\ 10^{\frac{dB}{20}} &=& \frac{p_1}{p_2} \qquad &| \qquad \cdot p_2 \\\\ p_2\cdot 10^{\frac{dB}{20}} &=& p_1\\\\ \boxed{~ \mathbf{ p_1 } \mathbf{=} \mathbf{p_2\cdot 10^{\frac{dB}{20}} } ~}\\\\ \end{array}$

a/(a+b)+a*a/(a*a-b*b)+b/(a-b)=?

$\begin{array}{lcl} \frac{a}{a+b} +a\cdot \left( \frac{a}{a\cdot a-b\cdot b} \right) +\frac{b}{a-b}&& \\\\ =\frac{a}{a+b} +\frac{a\cdot a}{a\cdot a-b\cdot b} +\frac{b}{a-b} \\\\ =\frac{a}{a+b} +\frac{a^2}{a^2-b^2} +\frac{b}{a-b} \qquad | \qquad \text{3. Binom }~ a^2-b^2 = (a+b)(a-b)\\\\ =\frac{a}{a+b} +\frac{a^2}{(a+b)(a-b)} +\frac{b}{a-b} \\\\ =\frac{a}{a+b} +\frac{a}{a+b} \cdot \frac{a}{a-b} +\frac{b}{a-b} \\\\ =\frac{a}{a+b} \left( 1+ \frac{a}{a-b} \right) +\frac{b}{a-b} \\\\ =\frac{a}{a+b} \left( \frac{a-b+a}{a-b} \right) +\frac{b}{a-b} \\\\ =\frac{a}{a+b} \left( \frac{2a-b}{a-b} \right) +\frac{b}{a-b} \\\\ =\frac{1}{a-b} \left( \frac{a(2a-b)}{a+b} +b \right) \\\\ =\frac{1}{a-b} \left( \frac{a(2a-b)+b(a+b)}{a+b} \right) \\\\ =\frac{a(2a-b)+b(a+b)}{(a-b)(a+b)}\\\\ =\frac{2a^2-ab+ba+b^2}{(a-b)(a+b)} \qquad | \qquad -ab+ba = 0\\\\ =\frac{2a^2+b^2}{(a-b)(a+b)} \qquad | \qquad \text{3. Binom }~ (a+b)(a-b)= a^2-b^2 \\\\ \mathbf{=\frac{2a^2+b^2}{a^2-b^2 }} \\\\ \mathbf{\frac{a}{a+b} +a\cdot \left( \frac{a}{a\cdot a-b\cdot b} \right) +\frac{b}{a-b}=\frac{2a^2+b^2}{a^2-b^2 }} \\\\ \end{array}$

Ich soll eine gleichung 4. Grades aufstellen mit folgenden Lösungen: {nl} x1= -2+i {nl} x2= 8+3i {nl} x3 =-2-i {nl} x4 = 8-3i

$\begin{array}{lcll} ( x+2-i)(x-8-3i)\cdot\\ (x+2+i)(x-8+3i)\\ &=&( x+2-i)(x+2+i)\cdot (x-8-3i)(x-8+3i) \qquad & | \qquad \text{3. Binom}\\ &=&[ (x+2)^2-i^2]\cdot [(x-8)^2-(3i)^2] \qquad & | \qquad i^2 = -1\\ &=&[ (x+2)^2+1]\cdot [(x-8)^2+9] \\ &=&( x^2+4x+4+1)\cdot (x^2-16x+64+9)\\ &=&( x^2+4x+5)\cdot (x^2-16x+73)\\ &=&x^4-16x^3+73x^2+4x^3-64x^2+292x+5x^2-80x+365\\ &=&x^4-12x^3+14x^2+212x+365 \end{array}$

Part 2:

$\begin{array}{rcl} \lim \limits_{x\to 0} { \left( \frac{ \sin{(x)} } {x} \right)^{ \cot{(x)} } }=\ ?\\ \left( \frac{ \sin{(x)} } {x} \right)^{ \cot{(x)} } = e^{ \ln{ \left( \frac{ \sin{(x)} } {x} \right) \cdot \cot{(x)} } } = e^{ \ln{ \left( 1+ \frac{\sin{(x)}-x}{x} \right) } \cdot \cot{(x)} }\\\\ \ln{ \left( 1+ \frac{\sin{(x)}-x}{x} \right) } \cdot \cot{(x)} = \left[ \left( -\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^1 \\ -\frac12 \left(-\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^2\\ +\frac13 \left(-\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^3\\ -\frac14 \left(-\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^4 \right] \cdot \\ \left( \frac{1}{x} -\frac13 x -\frac{1}{45} x^3 -\frac{2}{945} x^5 -\frac{1}{4725} x^7 -\cdots \right) \end{array}$

$\begin{array}{rcl} \ln{ \left( 1+ \frac{\sin{(x)}-x}{x} \right) } \cdot \cot{(x)} = \left[ \left( -\frac{x^2}{6} +\frac{x^4}{120} -\frac{x^6}{5040} +\frac{x^8}{40320} -+\cdots \right)\\ -\frac12 \left(\frac{x^4}{36} -\frac{x^6}{360} +\frac{41 x^8}{302400} -+\cdots \right)\\ +\frac13 \left(-\frac{x^6}{216} +\frac{x^8}{1440} -+\cdots \right)\\ -\frac14 \left(\frac{x^8}{1296} -+\cdots \right) \right] \cdot \\ \left( \frac{1}{x} -\frac13 x -\frac{1}{45} x^3 -\frac{2}{945} x^5 -\frac{1}{4725} x^7 -\cdots \right) \end{array}$

until $\mathbf{x^7}$:

$\begin{array}{llcl} \ln{ \left( 1+ \frac{\sin{(x)}-x}{x} \right) } \cdot \cot{(x)} = \\ \mathbf{ \times (\frac{1}{x} ): } & -\frac{x}{6} +\frac{x^3}{120} -\frac{x^5}{5040} +\frac{x^7}{40320} -+\cdots \\ &-\frac{x^3}{72} +\frac{x^5}{720} -\frac{41x^7}{604800} +-\cdots\\ &-\frac{x^5}{648} +\frac{x^7}{4320} -+\cdots -\frac{x^7}{5184} +-\cdots\\ \mathbf{ \times (-\frac{1x}{3} ): } & +\frac{x^3}{18} -\frac{x^5}{360} +\frac{x^7}{15120} -+\cdots +\frac{x^5}{216} -\frac{x^7}{2160} +-\cdots +\frac{x^7}{1944} -+\cdots\\ \mathbf{ \times (-\frac{1x^3}{45} ): } & +\frac{x^5}{270} -\frac{x^7}{5400} -+\cdots +\frac{x^7}{3240} -+\cdots\\ \mathbf{ \times (-\frac{2x^5}{945} ): } & +\frac{x^7}{2835} -+\cdots \end{array}$

$\begin{array}{llcl} \ln{ \left( 1+ \frac{\sin{(x)}-x}{x} \right) } \cdot \cot{(x)} = \\ -\frac{x}{6}\\ +x^3(\frac{1}{120}-\frac{1}{72}+\frac{1}{18})\\ +x^5(-\frac{1}{5040}+\frac{1}{720}-\frac{1}{648}-\frac{1}{360}+\frac{1}{216}+\frac{1}{270})\\ +x^7(\frac{1}{40320}-\frac{41}{604800}+\frac{1}{4320}-\frac{1}{5184}+\frac{1}{15120}-\frac{1}{2160}+\frac{1}{1944}-\frac{1}{5400}+\frac{1}{3240}+\frac{1}{2835})\\ +\cdots\\\\ \boxed{~\ln{ \left( 1+ \frac{\sin{(x)}-x}{x} \right) } \cdot \cot{(x)} = -\frac{x}{6}+\frac{x^3}{20} +\frac{59x^5}{11340}+\frac{401x^7}{680400} +\cdots ~}\\\\ \lim \limits_{x\to 0} { \left( \frac{ \sin{(x)} } {x} \right)^{ \cot{(x)} } }\\ = \lim \limits_{x\to 0} { e^{ \ln{ \left( \frac{ \sin{(x)} } {x} \right) \cdot \cot{(x)} } } }\\ = \lim \limits_{x\to 0} { e^{ \ln{ \left( 1+ \frac{\sin{(x)}-x}{x} \right) } \cdot \cot{(x)} } }\\ = \lim \limits_{x\to 0} { e^{-\frac{x}{6}+\frac{x^3}{20} +\frac{59x^5}{11340}+\frac{401x^7}{680400} +\cdots } } =e^0\\ \boxed{~ \lim \limits_{x\to 0} { \left( \frac{ \sin{(x)} } {x} \right)^{ \cot{(x)} } } =1 ~} \end{array}$

ready.

$\lim \limits_{x\to 0} { \left( \frac{ \sin{(x)} } {x} \right)^{ \cot{(x)} } }= \ ?\\ $

Part 1:

We need

$\boxed{~ \begin{array}{rcl} \cot{(x)} &=& \frac{1}{x} -\frac13 x -\frac{1}{45} x^3 -\frac{2}{945} x^5 -\frac{1}{4725} x^7 -\cdots \\ \ln{(1+x)} &=& x -\frac{x^2}{2} +\frac{x^3}{3} -\frac{x^4}{4} +-\cdots \\ \hline \sin{(x)} &=& x -\frac{x^3}{3!} +\frac{x^5}{5!} -\frac{x^7}{7!} +\frac{x^9}{9!} -+\cdots \\ \sin{(x)}-x &=& -\frac{x^3}{3!} +\frac{x^5}{5!} -\frac{x^7}{7!} +\frac{x^9}{9!} -+\cdots \\ \frac{\sin{(x)}-x}{x} &=& -\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \\ \hline \frac{\sin{(x)}}{x} &=& 1+ \left( \frac{\sin{(x)}}{x} - 1 \right) = 1+ \frac{\sin{(x)}-x}{x} \end{array} ~}$

$\boxed{~ \begin{array}{rcl} \ln{(1+z)} &=& z -\frac{z^2}{2} +\frac{z^3}{3} -\frac{z^4}{4} +-\cdots \\ z &=& \frac{\sin{(x)}-x}{x} = -\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \\ \ln{ ( 1+ \frac{ \sin{(x)}-x } {x} ) } &=& \left( -\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^1 \\ &&-\frac12 \left(-\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^2\\ &&+\frac13 \left(-\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^3\\ &&-\frac14 \left(-\frac{x^2}{3!} +\frac{x^4}{5!} -\frac{x^6}{7!} +\frac{x^8}{9!} -+\cdots \right)^4 \end{array} ~}$

continued...

Find minimum and maximum values of the functioin f(x) = (3-x)**2 + (1+x)**3 for 0<=x<=2

$\begin{array}{rcl} f(x) &=& (3-x)^2 + (1+x)^3 \\\\ f'(x) &=& 2\cdot (3-x)\cdot (-1) + 3\cdot (1+x)^2\cdot (1)\\ f'(x) &=& 3x^2+8x-3\\\\ f''(x) &=& 3\cdot 2\cdot x + 8 \\ f''(x) &=& 6x+8 \end{array}$

minimum and maximum: $f'(x) = 0$

$\begin{array}{rcl} 3x^2+8x-3 &=& 0 \\\\ \boxed{~ x = {-b \pm \sqrt{b^2-4ac} \over 2a} ~}\\ x_{1,2} &=& {-8 \pm \sqrt{8^2-4\cdot 3\cdot (-3) } \over 2\cdot 3} \\ x_{1,2} &=& {-8 \pm \sqrt{64+36} \over 6 } \\ x_{1,2} &=& {-8 \pm \sqrt{100} \over 6 } \\ x_{1,2} &=& {-8 \pm 10 \over 6 } \\\\ x_1 &=& {-8 + 10 \over 6 }\\ x_1 &=& {2 \over 6 }\\ \mathbf{x_1} &\mathbf{=}& \mathbf{\frac13} \qquad y''(\frac13) = 6\cdot \frac13 + 8 = 10 \quad \mathbf{>0} \quad \rightarrow \text{Minimum}\\\\ x_2 &=& {-8 - 10 \over 6 }\\ x_2 &=& {-18 \over 6 }\\ \mathbf{x_2} &\mathbf{=}& \mathbf{-3} \qquad y''(\frac13) = 6\cdot (-3) + 8 = -10 \quad \mathbf{<0} \quad \rightarrow \text{Maximum}\\\\\\ \end{array}$

$0 \le x \le 2$ there is a Minimum at $\mathbf{x = \frac13}$

Wie kann den Bruch 1/3 auf 1 bringen ?

$\frac13 \cdot \frac{1}{ \frac13 } = 1$

Sorry,

Bitte "kleiner" und "kleiner gleich" durch "größer" und "größer gleich" tauschen! Nur außerhalb der Umrahmungen.

Vielen Dank Gast!

Schade, ich kann leider meinen  Beitrag nicht mehr nachträglich ändern.

What is the units digit of 3^1992?

$3^{1992} \pmod {10} = \ ?\\$

$\begin{array}{rcl} 3^{4} \equiv 81 \equiv 1 \pmod {10} \end{array}$

$\begin{array}{rcl} 3^{1992} &\equiv& 3^{4\cdot 498}\pmod {10} \\ &\equiv& (3^{4})^{498}\pmod {10} \qquad 3^{4} \equiv 1 \pmod {10}\\ &\equiv& (1)^{498}\pmod {10}\\ &\equiv& 1 \pmod {10}\\ \mathbf{3^{1992}} & \mathbf{\equiv} & \mathbf{ 1\pmod {10} } \end{array}$