All Answers 357312 Answers

When I write the actual proof, I will modify it so it only does stuff to LHS

Will be gone for ~1.5 hours

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line 3 is not wrong

Its $2\cdot(\sin 2^{\circ}+\sin 4^{\circ}+\cdots+\sin 88^{\circ} +\sin 90^{\circ})=\frac{\cos 1^{\circ}}{\sin 1^{\circ}}$

NVM im a idiot

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!

Umm not sure.

I have seen that done convincingly before but it is not usual.

I do not think it is worded as a 'proof' but maybe just as 'show'.

You have not used either word in your original question.

I am not familiar with the  sum-to-product  identity.

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If I show that the equation simplifies to cos(1)=cos(1) or something, doesn't that work too?

Well then I do not think you should be working on both sides at once.

You work on one side and show that it is equivalent to the other side.

Please do not delete your questions.  Someone could have been working on it.

Just make a new post saying that you have your answer, but also thank anyone who had already donated any of their time.

You could even give the final answer or an outline of how you solved it.

Maybe someone else will learn from you !

I'm trying to prove that the average of n $n \sin n^\circ$($n = 2, 4, 6, \ldots, 180$) is $\cot 1^\circ$

What are you actually trying to do?

I have not checked the last three lines but the rest of the logic looks ok.

BUT what is the question?

Thanks, asinus!

OK, thanks!

Ok thanks for telling me

An Amazing and Very COOL Presentation, Heureka! 

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}$

$a) ~C(x)=24000+80x\\~\\ b)~R(x)=320x\\~\\ c)~C(x)=R(x)\\ 24000+80x=320x\\ 24000=240x\\ x = 100$

I don't think any normal users can delete questions- I think only Mods or Admins can...

But I think its 50. Because you should count every angle, and as long as they have different names, I think it counts. Not sure...

No...we have  25 pairs  (by my count)  ....for example CKJ and IKD  is the same pair as IKD and CKJ

To have one real solution the discriminant must = 0   ....so...

(2b)^2 - 4 (1) (a - b)  = 0

4b^2 - 4a + 4b  =  0

b^2 - a + b  = 0

b^2 + b  =   a

By the above computer code.

$\text{one real solution means that the discriminant is zero}\\ \text{in the quadratic equation $a x^2 + b x +c$ the discriminant is given by $D=b^2-4ac$}\\ \text{here}\\ a=1,~b=2b,~c = a-b\\ D=(2b)^2 - 4(a-b) = 4b^2 - 4a+4ab\\~\\ D=0 \Rightarrow\\ 4b^2-4a+4ab=0\\ a = b+b^2$

so isnt it 50?

An easier - and more straightforward -  way to do this

log _x   32  = 5/2

This says that

x^(5/2)  = 32       take each side to the (2/5) power

x  = 32^(2/5)  =  [ 32^(1/5) } ^2   =  [ 2]^2   = 4