heureka

cos^2(pi/8)+cos^2(3pi/8)+cos^2(5pi/8)+cos^2(7pi/8)
can you solve it showing me the whole way of how doing it?

\(\begin{array}{l} \cos^2(\frac18\pi)+\cos^2(\frac38\pi)+\cos^2(\frac58\pi)+\cos^2(\frac78\pi) = ? \\ \end{array}\)

\(\begin{array}{|lcll|} \hline \cos(\frac38\pi) = \sin(\frac12\pi-\frac38\pi) = \sin(\frac18\pi) \\ \cos(\frac58\pi) = \sin(\frac12\pi-\frac58\pi) = -\sin(\frac18\pi) = -\sin(\pi -\frac18\pi ) = -\sin(\frac78 \pi) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \cos^2(\frac18\pi)+\cos^2(\frac38\pi)+\cos^2(\frac58\pi)+\cos^2(\frac78\pi) \\ &=& \cos^2(\frac18\pi)+ \Big[ \sin(\frac18\pi) \Big]^2 + \Big[ -\sin(\frac78 \pi) \Big]^2 +\cos^2(\frac78\pi) \\ &=& \underbrace{ \cos^2(\frac18\pi)+ \sin^2(\frac18\pi) }_{=1} + \underbrace{ \sin^2(\frac78 \pi) +\cos^2(\frac78\pi) }_{=1} \\ &=& 1+1 \\ &=& 2 \\ \hline \end{array} \)

cos^2(pi/8)+cos^2(3pi/8)+cos^2(5pi/8)+cos^2(7pi/8)

can you solve it showing me the whole way of how doing it?

\(\begin{array}{l} \cos^2(\frac18\pi)+\cos^2(\frac38\pi)+\cos^2(\frac58\pi)+\cos^2(\frac78\pi) =\ ? \\ \end{array}\)

\(\begin{array}{|rcll|} \hline \cos(\frac78\pi) = \cos(\pi - \frac18\pi) = -\cos(\frac18\pi) \\ \cos(\frac58\pi) = \cos(\pi - \frac38\pi) = -\cos(\frac38\pi) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \cos^2(\frac18\pi)+\cos^2(\frac38\pi)+\cos^2(\frac58\pi)+\cos^2(\frac78\pi) \\ &=& \cos^2(\frac18\pi)+\cos^2(\frac38\pi) + \Big[ -\cos(\frac38\pi) \Big]^2 + \Big[ -\cos(\frac18\pi) \Big]^2 \\ &=& \cos^2(\frac18\pi)+\cos^2(\frac38\pi) + \cos^2(\frac38\pi) + \cos^2(\frac18\pi) \\ &=& 2\cdot \Big( \cos^2(\frac18\pi)+\cos^2(\frac38\pi) \Big) \quad | \quad \cos(\frac38\pi) = \sin(\frac12\pi-\frac38\pi) = \sin(\frac18\pi) \\ &=& 2\cdot \Big( \underbrace{ \cos^2(\frac18\pi)+\sin^2(\frac18\pi) }_{=1} \Big) \\ &=& 2 \\ \hline \end{array}\)

which whole numbers between 2 and 40 can be divided by 4 with a remainder of 2

and also be divided by 5 with a remainder of 1

\(\begin{array}{rcll} n &\equiv& {\color{red}2} \pmod {{\color{green}4}} \\ n &\equiv& {\color{red}1} \pmod {{\color{green}5}} \\ \text{Let } m &=& 4\cdot 5 = 20 \\ \end{array} \)

Because 4 and 5 are relatively prim ( gcd(4,5) = 1 ) we can go on:

\(\begin{array}{rcll} x &=& {\color{red}2} \cdot {\color{green}5} \cdot [ \frac{1}{ {\color{green}5} } \pmod{{\color{green}4}} ] +{\color{red}1} \cdot {\color{green}4} \cdot [ \frac{1}{ {\color{green}4} } \pmod{{\color{green}5}} ] +4\cdot 5 \cdot n \quad & | \quad n \in Z \\\\ && [ \frac{1}{ {\color{green}5} } \pmod{{\color{green}4}} ] \\ &=& [ {\color{green}5}^{\varphi({\color{green}4})-1} \pmod {{\color{green}4}} ] \quad & | \quad \varphi({\color{green}4})=4\cdot(1-\frac12)=2 \\ &=& [ {\color{green}5}^{2-1} \pmod {{\color{green}4}} ] \\ &=& [ {\color{green}5}^{1} \pmod {{\color{green}4}} ] \\ &=& [ {\color{green}1} \pmod {{\color{green}4}} ] \\ &=& [ 1] \\\\ && [ \frac{1}{ {\color{green}4} } \pmod{{\color{green}5}} ] \\ &=& [ {\color{green}4}^{\varphi({\color{green}5})-1} \pmod {{\color{green}5}} ] \quad & | \quad \varphi({\color{green}5})=5\cdot(1-\frac15)=4 \\ &=& [ {\color{green}4}^{4-1} \pmod {{\color{green}5}} ] \\ &=& [ {\color{green}4}^{3} \pmod {{\color{green}5}} ] \\ &=& [ {\color{green}4} \pmod {{\color{green}5}} ] \\ &=& [ 4 ] \\\\ x &=& {\color{red}2} \cdot {\color{green}5} \cdot [ 1] + {\color{red}1} \cdot {\color{green}4} \cdot [4] +4\cdot 5 \cdot n \quad & | \quad n \in Z \\\\ x &=& 10+16+20\cdot n \\ x &=& 26 + 20\cdot n \quad & | \quad x_{\text{min}} = 26 \pmod {m} \\ && & | \quad = 26 \pmod {20} = 6 \pmod {20} \\ x &=& 6 + 20\cdot n \quad & | \quad n \in Z \\\\ x_1 &=& 6 + 20\cdot 0 \quad & | \quad n=0 \\ x_1 &=& 6\checkmark \quad & | \quad 2\le x\le40 \\\\ x_2 &=& 6 + 20\cdot 1 \quad & | \quad n=1 \\ x_2 &=& 26\checkmark \quad & | \quad 2\le x\le40 \\ \end{array}\)

Kann die Gleichung jemand nach F1 auflösen ?

a*ln(f1)+b*ln((f1*w1)/(a*w2)*b)=x

\(\begin{array}{|rcll|} \hline a\cdot \ln(f_1)+b\cdot \ln \Big( \frac{f_1\cdot w_1} {a\cdot w_2} \cdot b \Big) &=& x \\ a\cdot \ln(f_1)+b\cdot \ln \Big( f_1 \cdot \frac{b\cdot w_1} {a\cdot w_2} \Big) &=& x \\ a\cdot \ln(f_1)+b\cdot \Big[ \ln( f_1 ) + \ln \Big( \frac{b\cdot w_1} {a\cdot w_2} \Big) \Big] &=& x \\ a\cdot \ln(f_1)+b\cdot \ln( f_1 ) + b\cdot \ln \Big( \frac{b\cdot w_1} {a\cdot w_2} \Big) &=& x \\ (a+b) \cdot \ln(f_1) + b\cdot \ln \Big( \frac{b\cdot w_1} {a\cdot w_2} \Big) &=& x \quad & | \quad - b\cdot \ln \Big( \frac{b\cdot w_1} {a\cdot w_2} \Big) \\ (a+b) \cdot \ln(f_1) &=& x - b\cdot \ln \Big( \frac{b\cdot w_1} {a\cdot w_2} \Big) \quad & | \quad : (a+b) \\ \ln(f_1) &=& \dfrac{ x - b\cdot \ln \Big( \frac{b\cdot w_1} {a\cdot w_2} \Big) } {a+b} \\ e^{\ln(f_1)} &=& e^{\Big( \dfrac{ x - b\cdot \ln ( \frac{b\cdot w_1} {a\cdot w_2} ) } {a+b} \Big) } \\ \mathbf{ f_1 } & \mathbf{=} & \mathbf{ e^{\Big( \dfrac{ x - b\cdot \ln ( \frac{b\cdot w_1} {a\cdot w_2} ) } {a+b} \Big) } } \\ \hline \end{array}\)

the angles of a triangle are in the ratio 3:5:4 calculate the size of each angle

\(\begin{array}{rcll} \text{Let } \alpha &=& \text{angle}_1 \\ \text{Let } \beta &=& \text{angle}_2 \\ \text{Let } \gamma &=& \text{angle}_3 \\ \gamma &=& 180^{\circ}-(\alpha+\beta) \\ \end{array} \)

\(\begin{array}{|rcll|} \hline && \alpha : \beta : \gamma \\ &=& \alpha : \beta : 180^{\circ}-(\alpha+\beta) \\ &=& 3:5:4 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1): & \frac{\alpha}{\beta} &=& \frac{3}{5} \\ & \alpha &=& \frac{3}{5}\cdot \beta \\\\ (2): & \frac{\alpha}{\gamma} = \frac{\alpha} { 180^{\circ}-(\alpha+\beta) } &=& \frac{3}{4} \\ & \frac{\alpha} { 180^{\circ}-(\alpha+\beta) } &=& \frac{3}{4} \\ & \frac{4}{3} \cdot \alpha &=& 180^{\circ}-(\alpha+\beta) \quad & | \quad \alpha = \frac{3}{5}\cdot \beta \\ & \frac{4}{3} \cdot \frac{3}{5}\cdot \beta &=& 180^{\circ}-(\frac{3}{5}\cdot \beta +\beta) \\ & \frac{4}{5} \cdot \beta &=& 180^{\circ}-\frac{8}{5}\cdot \beta \\ & \frac{4}{5} \cdot \beta +\frac{8}{5}\cdot \beta &=& 180^{\circ} \\ & \frac{12}{5} \cdot \beta &=& 180^{\circ} \\ & \beta &=& 180^{\circ}\cdot \frac{5}{12} \\ & \mathbf{ \beta } & \mathbf{=} & \mathbf{ 75^{\circ} } \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline & \alpha &=& \frac{3}{5}\cdot \beta \quad & | \quad \mathbf{ \beta = 75^{\circ} } \\ & \alpha &=& \frac{3}{5}\cdot 75^{\circ} \\ & \mathbf{ \alpha } & \mathbf{=} & \mathbf{ 45^{\circ} } \\ \\ & \gamma &=& 180^{\circ}-(\alpha+\beta) \quad & | \quad \mathbf{ \alpha = 45^{\circ} } \qquad \mathbf{ \beta = 75^{\circ} } \\ & \gamma &=& 180^{\circ}-( 45^{\circ}+75^{\circ} ) \\ & \gamma &=& 180^{\circ}-120^{\circ} \\ & \mathbf{ \gamma } & \mathbf{=} & \mathbf{ 60^{\circ} } \\ \hline \end{array} \)

The angles of the triangle are \(45^{\circ},\ 75^{\circ},\ 60^{\circ}\)

12 1/2 + 4 2/5 - 8 3/4 =

1.

\(\begin{array}{|rcll|} \hline && 12 \frac12 + 4 \frac25 - 8 \frac34 \\ &=& 12+4-8 + \frac12 + \frac25 - \frac34 \\ &=& 8 + 0.5 + 0.4 - 0.75 \\ &=& 8 + 0.15 \\ &=& 8.15 \\ &=& 8 \frac{15}{100} \\ &=& 8 \frac{3}{20} \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline && 12 \frac12 + 4 \frac25 - 8 \frac34 \\ &=& 12+4-8 + \frac12 + \frac25 - \frac34 \\ &=& 8 + \frac12\cdot \frac{10}{10} + \frac25\cdot \frac{4}{4} - \frac34\cdot \frac{5}{5} \\ &=& 8 + \frac{10}{20} + \frac{8}{20} - \frac{15}{20} \\ &=& 8 + \frac{10+8-15}{20} \\ &=& 8 \frac{3}{20} \\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline && 12 \frac12 + 4 \frac25 - 8 \frac34 \\ &=& \frac{2\cdot 12+1}{2} + \frac{4\cdot 5+2}{5} + \frac{4\cdot 8+3}{4} \\ &=& \frac{25}{2} + \frac{22}{5} + \frac{35}{4} \\ &=& \frac{25}{2}\cdot \frac{10}{10} + \frac{22}{5}\cdot \frac{4}{4} - \frac{35}{4}\cdot \frac{5}{5} \\ &=& \frac{250}{20} + \frac{88}{20} - \frac{175}{20} \\ &=& \frac{250+88-175}{20} \\ &=& \frac{163}{20} \\ &=& 8 \frac{3}{20} \\ \hline \end{array}\)

find the area of a triangle with A=29 degrees b=49 and c=50

Formula:

\(\begin{array}{|rcll|} \hline \text{Area} &=& \frac12 \cdot b\cdot c\cdot \sin(A) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{Area} &=& \frac12 \cdot 49\cdot 50\cdot \sin(29^{\circ}) \\ &=& \frac12 \cdot 49\cdot 50\cdot 0.48480962025 \\ &=& 593.891784802 \\ \hline \end{array} \)

what is the mathematical function mod

Formula:

\(a \text{ mod } b = a-b\cdot [\frac ab] \qquad \qquad [\frac ab] \text{ is the integer part of a divided b}\)

Example:

\(102 \text{ mod } 12 = 102-12\cdot [\frac{102}{12}] \\ 102 \text{ mod } 12 = 102-12\cdot [8.5] \\ 102 \text{ mod } 12 = 102-12\cdot 8 \\ 102 \text{ mod } 12 = 102-96 \\ 102 \text{ mod } 12 = 6 \)

The remainder of \(\frac{102}{12} \) is 6

If |2x – 4| = 8 then x = _____ and _____

Formula:

\(\begin{array}{|rcll|} \hline x^2=|x^2| = |x|^2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline |2x – 4| &=& 8 \\ |2x – 4|^2 &=& 8^2 \quad & |2x – 4|^2 = (2x – 4)^2 \\ (2x – 4)^2 &=& 8^2 \\ [2\cdot(x – 2)]^2 &=& 8^2 \\ 2^2\cdot(x – 2)^2 &=& 8^2 \\ 4\cdot (x – 2)^2 &=& 64 \quad & | \quad : 4 \\ (x – 2)^2 &=& 16 \quad & | \quad \text{square root both sides } \\ x – 2 &=& \pm\sqrt{16} \\ x – 2 &=& \pm 4 \quad & | \quad +2 \\ x &=& 2\pm 4 \\\\ x_1 &=& 2+4 \\ x_1 &=& 6 \\\\ x_2 &=& 2-4 \\ x_2 &=& -2 \\ \hline \end{array} \)

x = -2 and  6

wenn ich ein Lampen System habe mit drei Lampen die gerade alle aus sind

wie viele mögliche Stellungen kann ich dann einstellen

in dem ich Lampen an und oder aus mache?

\(\begin{array}{|r|r|r|r|} \hline & \text{Lampe } 1 & \text{Lampe } 2 & \text{Lampe } 3 \\ \hline 1 & \color{red}{\text{aus}} & \color{red}{\text{aus}}& \color{red}{\text{aus}} \\ 2 & \color{red}{\text{aus}} & \color{red}{\text{aus}} & \color{green}{\text{an}} \\ 3 & \color{red}{\text{aus}} & \color{green}{\text{an}} & \color{red}{\text{aus}} \\ 4 & \color{red}{\text{aus}} & \color{green}{\text{an}} & \color{green}{\text{an}} \\ 5 & \color{green}{\text{an}} & \color{red}{\text{aus}} & \color{red}{\text{aus}} \\ 6 & \color{green}{\text{an}} & \color{red}{\text{aus}} & \color{green}{\text{an}} \\ 7 & \color{green}{\text{an}} & \color{green}{\text{an}} & \color{red}{\text{aus}} \\ 8 & \color{green}{\text{an}} & \color{green}{\text{an}} & \color{green}{\text{an}} \\ \hline \end{array}\)