heureka

if line 3 is not wrong, why don't you get  \(\cos 1^{\circ} = \cos 1^{\circ}\) ?

I got it out !

What did I do wrong?

\(2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} = 90\cdot \cot 1^{\circ}\)

\(\small{ \begin{array}{|rcll|} \hline \mathbf{2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 176 \sin 176^{\circ} + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} } &=& \mathbf{90\cdot \cot 1^{\circ}} \\\\ 2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 88 \sin 88^{\circ}+ 90 \sin 90^{\circ} + 92 \sin 92^{\circ} + \cdots + 176 \sin 176^{\circ}+178 \sin 178^{\circ} + 180 \underbrace{\sin 180^{\circ}}_{=0} &=& 90\cdot \cot 1^{\circ} \\ 2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 88 \sin 88^{\circ}+ 90 \sin 90^{\circ} + 92 \sin 92^{\circ} + \cdots + 176 \sin 176^{\circ}+ 178 \sin 178^{\circ} &=& 90\cdot \cot 1^{\circ} \\ 2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 88 \sin 88^{\circ}+ 90 \sin 90^{\circ} + 92 \sin (180^\circ -92^{\circ}) + \cdots + 176 \sin (180^\circ-176^{\circ} ) + 178 \sin (180^\circ-178^{\circ} ) &=& 90\cdot \cot 1^{\circ} \\ 2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 88 \sin 88^{\circ}+ 90 \sin 90^{\circ} + 92 \sin 88^{\circ} + \cdots + 176 \sin 4^{\circ}+178 \sin 2^{\circ} &=& 90\cdot \cot 1^{\circ} \\ 2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 88 \sin 88^{\circ}+ 90 \sin 90^{\circ} \\ 178 \sin 2^{\circ} + 176 \sin 4^{\circ} + \cdots + 92 \sin 88^{\circ} &=& 90\cdot \cot 1^{\circ} \\ (2+178)\sin 2^{\circ}+(4+176)\sin 4^{\circ} + \cdots +(88+92)\sin 88^{\circ}+ 90 \sin 90^{\circ} &=& 90\cdot \cot 1^{\circ} \\ 180\sin 2^{\circ}+180\sin 4^{\circ} + \cdots +180\sin 88^{\circ}+ 90 \sin 90^{\circ} &=& 90\cdot \cot 1^{\circ} \quad | \quad : 90 \\ 2\sin 2^{\circ}+2\sin 4^{\circ} + \cdots +2\sin 88^{\circ}+ \sin 90^{\circ} &=& \cot 1^{\circ} \\ 2\sin 2^{\circ}+2\sin 4^{\circ} + \cdots +2\sin 88^{\circ}+ \sin 90^{\circ} &=& \dfrac{\cos 1^{\circ}}{\sin 1^{\circ}} \quad | \quad \cdot \sin 1^{\circ} \\ 2\sin 1^{\circ}\sin 2^{\circ}+2\sin 1^{\circ}\sin 4^{\circ} + \cdots +2\sin 1^{\circ}\sin 88^{\circ}+ \sin 1^{\circ}\sin 90^{\circ} &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\cos 3^{\circ}+ \cos 3^{\circ}-\cos 5^{\circ} + \cdots + \cos 87^{\circ}-\cos 89^{\circ}+ \sin 1^{\circ}\sin 90^{\circ} &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\cos 89^{\circ}+ \underbrace{ \sin 1^{\circ}\sin 90^{\circ}}_{=\frac12 (\cos 89^\circ - \cos 91^\circ )} &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\cos 89^{\circ}+ \frac12 (\cos 89^\circ - \cos 91^\circ ) &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\cos 89^{\circ}+ \frac12 \cos 89^\circ -\frac12 \cos 91^\circ &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\frac12 \cos 89^\circ -\frac12 \cos 91^\circ &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\frac12 (\cos 89^\circ + \cos 91^\circ) &=& \cos 1^{\circ} \\ \boxed{ \cos 91^\circ = -\cos(180^\circ -91^\circ) \\ = -\cos 89^\circ }\\ \cos 1^{\circ}-\frac12 (\cos 89^\circ -\cos 89^\circ ) &=& \cos 1^{\circ} \\ \cos 1^{\circ}-0 &=& \cos 1^{\circ} \\ \mathbf{ \cos 1^{\circ} } &=& \mathbf{ \cos 1^{\circ} } \\ \hline \end{array} } \)

What did I do wrong?

Line 3:  \(2\cdot(\sin 2^{\circ}+\sin 4^{\circ}+\cdots+\sin 88^{\circ} )+\sin 90^{\circ}=\frac{\cos 1^{\circ}}{\sin 1^{\circ}}\)

\(\sin 90^{\circ} \) only once!

How many five digit even integers sums up to 13

\(\begin{array}{|l|l|r|r|r|} \hline \text{5 digit even integers} & \text{partition} & \text{permutation} & - \text{partition} &- \text{permutation} \\ \hline 9\{4,0,0\}0 & P(4,1), P(4,2), P(4,3) & \\ &\{4,0,0\},\{3,1,0\},\{2,1,1\} & \binom{6}{2} \\ & \qquad\qquad \{2,2,0\} & \\ 9\{2,0,0\}2 & P(2,1), P(2,2), P(2,3) & \\ &\{2,0,0\},\{1,1,0\} & \binom{4}{2} \\ 9\{0,0,0\}4 & 1 & \frac{3!}{3!}=1= \binom{2}{2} \\ \hline 8\{5,0,0\}0 & P(5,1), P(5,2), P(5,3) & \\ &\{5,0,0\},\{4,1,0\},\{3,1,1\} & \binom{7}{2} \\ & \qquad\qquad \{3,2,0\},\{2,2,1\} & \\ 8\{3,0,0\}2 & P(3,1), P(3,2), P(3,3) & \\ &\{3,0,0\},\{2,1,0\},\{1,1,1\} & \binom{5}{2} \\ 8\{1,0,0\}4 &P(1,1), P(1,2), P(1,3) & \\ &\{1,0,0\} & \binom{3}{2} \\ \hline 7\{6,0,0\}0 & P(6,1), P(6,2), P(6,3) & \binom{8}{2} \\ 7\{4,0,0\}2 & P(4,1), P(4,2), P(4,3) & \binom{6}{2} \\ 7\{2,0,0\}4 & P(2,1), P(2,2), P(2,3) & \binom{4}{2} \\ 7\{0,0,0\}6 & 1 & \frac{3!}{3!}=1= \binom{2}{2} \\ \hline 6\{7,0,0\}0 & P(7,1), P(7,2), P(7,3) & \binom{9}{2} \\ 6\{5,0,0\}2 & P(5,1), P(5,2), P(5,3) & \binom{7}{2} \\ 6\{3,0,0\}4 & P(3,1), P(3,2), P(3,3) & \binom{5}{2} \\ 6\{1,0,0\}6 & P(1,1), P(1,2), P(1,3) & \binom{3}{2} \\ \hline 5\{8,0,0\}0 & P(8,1), P(8,2), P(8,3) & \binom{10}{2} \\ 5\{6,0,0\}2 & P(6,1), P(6,2), P(6,3) & \binom{8}{2} \\ 5\{4,0,0\}4 & P(4,1), P(4,2), P(4,3) & \binom{6}{2} \\ 5\{2,0,0\}6 & P(2,1), P(2,2), P(2,3) & \binom{4}{2} \\ 5\{0,0,0\}8 & 1 & \frac{3!}{3!}=1= \binom{2}{2} \\ \hline 4\{9,0,0\}2 & P(9,1), P(9,2), P(9,3) & \binom{11}{2} \\ 4\{7,0,0\}2 & P(7,1), P(7,2), P(7,3) & \binom{9}{2} \\ 4\{5,0,0\}4 & P(5,1), P(5,2), P(5,3) & \binom{7}{2} \\ 4\{3,0,0\}6 & P(3,1), P(3,2), P(3,3) & \binom{5}{2} \\ 4\{1,0,0\}8 & P(1,1), P(1,2), P(1,3) & \binom{3}{2} \\ \hline 3\{10,0,0\}0 & P(10,1), P(10,2), P(10,3) & \binom{12}{2} & \{10,0,0\} & - \frac{3!}{1!2!} \\ 3\{8,0,0\}2 & P(8,1), P(8,2), P(8,3) & \binom{10}{2} \\ 3\{6,0,0\}4 & P(6,1), P(6,2), P(6,3) & \binom{8}{2} \\ 3\{4,0,0\}6 & P(4,1), P(4,2), P(4,3) & \binom{6}{2} \\ 3\{2,0,0\}8 & P(2,1), P(2,2), P(2,3) & \binom{4}{2} \\ \hline 2\{11,0,0\}0 & P(11,1), P(11,2), P(11,3) & \binom{13}{2} & \{11,0,0\} & - \frac{3!}{1!2!} \\ & & & \{10,1,0\} & - \frac{3!}{1!1!1!} \\ 2\{9,0,0\}2 & P(9,1), P(9,2), P(9,3) & \binom{11}{2} \\ 2\{7,0,0\}4 & P(7,1), P(7,2), P(7,3) & \binom{9}{2} \\ 2\{5,0,0\}6 & P(5,1), P(5,2), P(5,3) & \binom{7}{2} \\ 2\{3,0,0\}8 & P(3,1), P(3,2), P(3,3) & \binom{5}{2} \\ \hline 1\{12,0,0\}0 & P(12,1), P(12,2), P(12,3) & \binom{14}{2} & \{12,0,0\} & - \frac{3!}{1!2!} \\ & & & \{11,1,0\} & - \frac{3!}{1!1!1!} \\ & & & \{10,2,0\} & - \frac{3!}{1!1!1!} \\ & & & \{10,1,1\} & - \frac{3!}{1!2!} \\ 1\{10,0,0\}2 & P(10,1), P(10,2), P(10,3) & \binom{12}{2} & \{10,0,0\} & - \frac{3!}{1!2!} \\ 1\{8,0,0\}4 & P(8,1), P(8,2), P(8,3) & \binom{10}{2} \\ 1\{6,0,0\}6 & P(6,1), P(6,2), P(6,3) & \binom{8}{2} \\ 1\{4,0,0\}8 & P(4,1), P(4,2), P(4,3) & \binom{6}{2} \\ \hline \end{array} \)

Sum off all permutations:

\(\begin{array}{|rcll|} \hline && \binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2} \quad &|\quad 9\ldots , \text{ and } 8\ldots \\ &+& \binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2} \quad &|\quad 7\ldots ,\ \text{ and } 6\ldots \\ &+& \binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2}+ \binom{10}{2}+ \binom{11}{2} \quad &|\quad 5\ldots ,\ \text{ and } 4\ldots \\ &+& \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2}+ \binom{10}{2}+ \binom{11}{2}+ \binom{12}{2}+ \binom{13}{2}- 2\times\frac{3!}{1!2!}- 1\times \frac{3!}{1!1!1!} \quad &|\quad 3\ldots ,\ \text{ and } 2\ldots \\ &+& \binom{6}{2} + \binom{8}{2} + \binom{10}{2} + \binom{12}{2} + \binom{14}{2} - 3\times\frac{3!}{1!2!}- 2\times \frac{3!}{1!1!1!} \quad &|\quad 1\ldots \\\\ &=& \underbrace{\binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2} }_{= \binom{8}{3}\text{( hockey stick identity)} } \quad &|\quad 9\ldots , \text{ and } 8\ldots \\ &+& \underbrace{\binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2}}_{=\binom{10}{3}\text{( hockey stick identity)} } \quad &|\quad 7\ldots ,\ \text{ and } 6\ldots \\ &+& \underbrace{\binom{2}{2} + \binom{3}{2}+ \binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2}+ \binom{10}{2}+ \binom{11}{2} }_{=\binom{12}{3}\text{( hockey stick identity)} } \quad &|\quad 5\ldots ,\ \text{ and } 4\ldots \\ &+& \underbrace{\binom{4}{2}+ \binom{5}{2}+ \binom{6}{2}+ \binom{7}{2}+ \binom{8}{2}+ \binom{9}{2}+ \binom{10}{2}+ \binom{11}{2}+ \binom{12}{2}+ \binom{13}{2} }_{=\binom{14}{3} -\binom{3}{2} -\binom{2}{2} \text{( hockey stick identity)} }- 2\times\frac{3!}{1!2!}- 1\times \frac{3!}{1!1!1!} \quad &|\quad 3\ldots ,\ \text{ and } 2\ldots \\ &+& \binom{6}{2} + \binom{8}{2} + \binom{10}{2} + \binom{12}{2} + \binom{14}{2} - 3\times\frac{3!}{1!2!}- 2\times \frac{3!}{1!1!1!} \quad &|\quad 1\ldots \\\\ &=& \binom{8}{3} + \binom{10}{3} + \binom{12}{3} + \binom{14}{3} -\underbrace{\left(\binom{2}{2} +\binom{3}{2}\right)}_{=\binom{4}{3}\text{( hockey stick identity)} } \\ &+& \binom{6}{2} + \binom{8}{2} + \binom{10}{2} + \binom{12}{2} + \binom{14}{2} - 5\times\frac{3!}{1!2!}- 3\times \frac{3!}{1!1!1!} \\\\ &=& \binom{8}{3} + \binom{10}{3} + \binom{12}{3} + \binom{14}{3} - \binom{4}{3} \\ &+& \binom{6}{2} + \binom{8}{2} + \binom{10}{2} + \binom{12}{2} + \binom{14}{2} - 5\times\frac{3!}{1!2!}- 3\times \frac{3!}{1!1!1!} \\\\ && \boxed{\binom{8}{2}+\binom{8}{3} = \binom{9}{3} \\ \binom{10}{2}+\binom{10}{3} = \binom{11}{3} \\ \binom{12}{2}+\binom{12}{3} = \binom{13}{3} \\ \binom{14}{2}+\binom{14}{3} = \binom{15}{3} } \\\\ &=& \mathbf{\binom{9}{3} + \binom{11}{3} + \binom{13}{3} + \binom{15}{3} - \binom{4}{3} + \binom{6}{2} - 5\times\frac{3!}{1!2!}- 3\times \frac{3!}{1!1!1!}} \\\\ &=& \binom{9}{3} + \binom{11}{3} + \binom{13}{3} + \binom{15}{3} - \binom{4}{3} + \binom{6}{2} - 5\times 3- 3\times 6 \\\\ &=& \binom{9}{3} + \binom{11}{3} + \binom{13}{3} + \binom{15}{3} - \binom{4}{3} + \binom{6}{2} -33 \\\\ &=& 84 + 165 + 286 + 455 - 4 + 15 -33 \\ &=& \mathbf{968} \\ \hline \end{array}\)

The expression \(y^2+10y+33\) can be written as a combination of a square of a binomial and an integer.
Find the integer.

\(\begin{array}{|rcll|} \hline && y^2+10y+33 \\ &=& (y+5)^2 -25 + 33 \\ &=& (y+5)^2 +8 \\ \hline \end{array}\)

The red parabola shown is the graph of the equation \(x = ay^2 + by + c\). 

Find \(a+b+c\).

The Vertex of the parabola is \(P(x=-3,\ y=1)\)

\(\begin{array}{|rcll|} \hline \mathbf{ay^2 + by + c} &=& \mathbf{x} \quad | \quad x=-3,\ y=1 \\\\ a*(1)^2 + b*(1) + c &=& -3 \\ \mathbf{a + b + c} &=& \mathbf{-3} \\ \hline \end{array}\)

Thank you CPhill !

For the graph of a certain quadratic y = ax^2 + bx + c,
the vertex of the parabola is (3,7) and one of the x-intercepts is (-2,0).
What is the x-coordinate of the other x-intercept?

\(\text{Let $x-vertex = x_v = 3 $} \\ \text{Let $ x-intercept_1 = x_1 = -2 $} \\ \text{Let $ x-intercept_2 = x_2 =\ ? $} \)

\(\begin{array}{|rcll|} \hline \dfrac{x_1+x_2}{2} &=& x_v \\\\ \dfrac{-2+x_2}{2} &=& 3 \quad | \quad \cdot 2 \\\\ -2+x_2 &=& 6 \quad | \quad + 2 \\\\ \mathbf{ x_2 } &=& \mathbf{8} \\ \hline \end{array} \)

The x-coordinate of the other x-intercept is \(\mathbf{ 8 }\)

Was sind \(2^4*2^3*2\) ?

\(\begin{array}{|rcll|} \hline && 2^4*2^3*2 \\ &=& 2^4*2^3*2^1 \\ &=& 2^{4+3+1} \\ &=& 2^{8} \\ \hline \end{array}\)

3.
The sequence \((a_n)\) is defined recursively by \(a_0=1,\ a_1=\sqrt[19]{2}\), and \(a_n=a_{n-1}a_{n-2}^2\) for \(n\geq 2\).
What is the smallest positive integer k such that the product \(a_1a_2\cdots a_k\) is an integer?

\(\begin{array}{|l|r|rcl|l|} \hline & k & \text{product } a_1a_2\cdots a_k \\ \hline a_2 = a_1a_0^2=a_1^1 & 2 & a_1a_2 &=& a_1^2 & =2^{\frac{2}{19}} & = 1.07569058622 \\ \hline a_3 = a_2a_1^2=a_1^1a_1^2=a_1^3 & 3 & a_1a_2a_3 &=& a_1^5 & =2^{\frac{5}{19}} & = 1.20010271958 \\ \hline a_4 = a_3a_2^2=a_1^3a_1^2=a_1^5& 4 & a_1a_2\cdots a_4&=& a_1^{10} & =2^{\frac{10}{19}} & = 1.49375896165 \\ \hline a_{5} = a_4a_3^2=a_1^{5}a_1^{6}=a_1^{11} & 5 & a_1a_2\cdots a_5 &=& a_1^{21} & =2^{\frac{21}{19}} & = 2.15138117244 \\ \hline a_{6} = a_5a_4^2=a_1^{11}a_1^{10}=a_1^{21} & 6 & a_1a_2\cdots a_6 &=& a_1^{42} & =2^{\frac{42}{19}} & = 4.62844094913 \\ \hline a_{7} = a_6a_5^2=a_1^{21}a_1^{33}=a_1^{43} & 7 & a_1a_2\cdots a_7 &=& a_1^{85} & =2^{\frac{85}{19}} & = 22.2184182818 \\ \hline a_{8} = a_7a_6^2=a_1^{43}a_1^{42}=a_1^{85} & 8 & a_1a_2\cdots a_8 &=& a_1^{170} & =2^{\frac{170}{19}} & = 493.658110947 \\ \hline a_{9} = a_8a_7^2=a_1^{85}a_1^{86}=a_1^{171} & 9 & a_1a_2\cdots a_9 &=& a_1^{341} & =2^{\frac{341}{19}} & = 252752.952805 \\ \hline a_{10} = a_9a_8^2=a_1^{171}a_1^{170}=a_1^{341} & 10 & a_1a_2\cdots a_{10} &=& a_1^{682} & =2^{\frac{682}{19}} & = 2^{35.8947368421} \\ \hline a_{11} = a_{10}a_{9}^2=a_1^{341}a_1^{342}=a_1^{683} & 11 & a_1a_2\cdots a_{11} &=& a_1^{1365} & =2^{\frac{1365}{19}} & = 2^{71.8421052632} \\ \hline a_{12} = a_{11}a_{10}^2=a_1^{683}a_1^{682}=a_1^{1365} & 12 & a_1a_2\cdots a_{12} &=& a_1^{2730} & =2^{\frac{2730}{19}} & = 2^{143.684210526} \\ \hline a_{13} = a_{12}a_{11}^2=a_1^{1365}a_1^{1366}=a_1^{2731} & 13 & a_1a_2\cdots a_{13} &=& a_1^{5461} & =2^{\frac{5461}{19}} & = 2^{287.421052632} \\ \hline a_{14} = a_{13}a_{12}^2=a_1^{2731}a_1^{2730}=a_1^{5461} & 14 & a_1a_2\cdots a_{14} &=& a_1^{10922} & =2^{\frac{10922}{19}} & = 2^{574.842105263} \\ \hline a_{15} = a_{14}a_{13}^2=a_1^{5461}a_1^{5462}=a_1^{10923} & 15 & a_1a_2\cdots a_{15} &=& a_1^{21845} & =2^{\frac{21845}{19}} & = 2^{1,149.73684211} \\ \hline a_{16} = a_{15}a_{14}^2=a_1^{10923}a_1^{10922}=a_1^{21845} & 16 & a_1a_2\cdots a_{16} &=& a_1^{43690} & =2^{\frac{43690}{19}} & = 2^{2,299.47368421} \\ \hline a_{17} = a_{16}a_{15}^2=a_1^{21845}a_1^{21846}=a_1^{43691} & \mathbf{17} & a_1a_2\cdots a_{17} &=& a_1^{87381} & =2^{\frac{87381}{19}} & \mathbf{= 2^{4599}}\\ & & && & & \text{the product $a_1a_2\cdots a_{17}$} \\ & & && & & \text{with $k=17$ is an integer} \\ \hline \end{array}\)

The smallest positive integer \(k\) such that the product \(a_1a_2\cdots a_k\) is an integer is \(\mathbf{17}\)