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avatar+507 

\(2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots  + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} = 90\cdot \cot 1^{\circ}\)
\(2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots  + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} = 90\cdot \frac{\cos 1}{\sin 1}\)
We can use the identity \(\sin x = \sin (180^{\circ} - x)\) and simplify to get:
\(2\cdot(\sin 2^{\circ}+\sin 4^{\circ}+\cdots+\sin 88^{\circ} +\sin 90^{\circ})=\frac{\cos 1^{\circ}}{\sin 1^{\circ}}\)
We can multiply by \(\sin1^{\circ}\) to get:
\(2\cdot(\sin 2^{\circ}\cdot\sin 1^{\circ}+\sin 4^{\circ}\cdot\sin 1^{\circ}+\cdots+\sin 88^{\circ}\cdot\sin 1^{\circ}+\sin 90^{\circ}\cdot\sin 1^{\circ})=\cos 1^{\circ}\)
Then, we can use the sum-to-product (or Prosthaphaeresis :)) identities to get:
\(\cos1^{\circ}-\cos3^{\circ}+\cos3^{\circ}-\cos5^{\circ}+\cdots+\cos87^{\circ}-\cos89^{\circ}+\cos89^{\circ}-\cos91^{\circ}=\cos 1^\circ\)
We can cancel the terms to get:
\(\cos1^{\circ}-\cos91^{\circ}=\cos1^{\circ}\)

 

 

 

My other proof that also has some error in it:

\(2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots  + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} = 90\cdot \cot 1^{\circ}\)
\(2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots  + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} = 90\cdot \frac{\cos 1}{\sin 1}\)
We can use the identity \(\sin x = \sin (180^{\circ} - x)\) and simplify to get:
\(2\cdot(\sin 2^{\circ}+\sin 4^{\circ}+\cdots+\sin 88^{\circ})+2=\frac{\cos 1^{\circ}}{\sin 1^{\circ}}\)
We can multiply by \(\sin1^{\circ}\) to get:
\(2\cdot(\sin 2^{\circ}\cdot\sin 1^{\circ}+\sin 4^{\circ}\cdot\sin 1^{\circ}+\cdots+\sin 88^{\circ}\cdot\sin 1^{\circ})+2\cdot\sin1^{\circ}=\cos 1^{\circ}\)
Then, we can use the sum-to-product (or Prosthaphaeresis :)) identities to get:
\(\cos1^{\circ}-\cos3^{\circ}+\cos3^{\circ}-\cos5^{\circ}+\cdots+\cos87^{\circ}-\cos89^{\circ}+2\cdot\sin1^{\circ}=\cos 1^\circ\)
We can cancel the terms to get:
\(\cos1^{\circ}-\cos89^{\circ}+2\cdot\sin1^{\circ}=\cos1^{\circ}\)
We can use the identity \(\sin x=\cos({x-90^{\circ}})\) to get
\(\cos1^{\circ}+sin1^{\circ}=\cos1^{\circ}\)

 Jul 16, 2019
edited by Davis  Jul 16, 2019
edited by Davis  Jul 16, 2019
edited by Davis  Jul 16, 2019
edited by Davis  Jul 16, 2019
edited by Davis  Jul 16, 2019
edited by Davis  Jul 16, 2019
 #1
avatar+105464 
+1

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?

 Jul 16, 2019
 #2
avatar+507 
+1

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

.
 Jul 16, 2019
edited by Davis  Jul 16, 2019
edited by Davis  Jul 16, 2019
 #3
avatar+105464 
+1

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.

Melody  Jul 16, 2019
 #4
avatar+507 
+1

?

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

Davis  Jul 16, 2019
 #5
avatar+105464 
+1

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.

Melody  Jul 16, 2019
 #9
avatar+507 
+1

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

Davis  Jul 16, 2019
 #6
avatar+23273 
+3

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!

 

laugh

 Jul 16, 2019
 #7
avatar+507 
0

?

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

Davis  Jul 16, 2019
edited by Davis  Jul 16, 2019
edited by Davis  Jul 17, 2019
 #10
avatar+23273 
+3

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} } \)

 

 

laugh

heureka  Jul 16, 2019
 #12
avatar+104660 
+2

Very nice, heureka !!!!

 

 

cool cool cool

CPhill  Jul 16, 2019
 #13
avatar+23273 
+2

Thank You, CPhill !

 

laugh

heureka  Jul 17, 2019
 #8
avatar+507 
+1

Will be gone for ~1.5 hours

 Jul 16, 2019
 #11
avatar+23273 
+3

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

 

I got it out !

 

laugh

 Jul 16, 2019
edited by heureka  Jul 16, 2019
 #14
avatar+507 
+2

Thanks!

Davis  Jul 17, 2019
 #15
avatar+507 
+4

I re-wrote the proof by myself to make sure I got it:

Prove of \(\frac{2\cdot\sin2^{\circ}+4\cdot\sin4^{\circ}+\cdots+178\cdot\sin178^{\circ}+180\cdot\sin180^{\circ}}{90} = \cot 1^{\circ}\):

L.H.S.= \(\frac{2\cdot\sin2^{\circ}+4\cdot\sin4^{\circ}+\cdots+178\cdot\sin178^{\circ}+180\cdot\sin180^{\circ}}{90} = \cot 1^{\circ}\)
 
We can use the identity \(\sin x = \sin (180^{\circ} - x)\) to get:

L.H.S.= \(\frac{2\cdot\sin2^{\circ}+178\cdot\sin178^{\circ}+180\cdot\sin180^{\circ}+4\cdot\sin4^{\circ}+\cdots}{90} = \cot 1^{\circ}\)

We can simplify to get:

L.H.S.= \(2\cdot(\sin 2^{\circ}+\sin 4^{\circ}+\cdots+\sin 88^{\circ})+\sin 90^{\circ}\)

We can multiply by \(\frac{\sin1^{\circ}}{\sin1^{\circ}}\) to get:

L.H.S.= \(\frac{2\cdot(\sin 2^{\circ}\cdot\sin 1^{\circ}+\sin 4^{\circ}\cdot\sin 1^{\circ}+\cdots+\sin 88^{\circ}\cdot\sin 1^{\circ})+\sin 90^{\circ}\cdot\sin 1^{\circ}}{\sin1^{\circ}}\)

Then, we can use the sum-to-product (or Prosthaphaeresis :)) identities to get:

L.H.S.= \(\frac{\cos1^{\circ}-\cos3^{\circ}+\cos3^{\circ}-\cos5^{\circ}+\cdots+\cos87^{\circ}-\cos89^{\circ}+\sin90^{\circ}\cdot\sin1^{\circ}}{\sin1^{\circ}}\)

We can cancel the terms to get:

L.H.S.= \(\frac{\cos1^{\circ}-\cos89^{\circ}+\frac{\cos89^{\circ}-\cos91^{\circ}}{2}}{\sin1^{\circ}}\)

We can simplify to get:

L.H.S.= \(\frac{\cos1^{\circ}-\frac{\cos89^{\circ}+\cos91^{\circ}}{2}}{\sin1^{\circ}}\)

We can use the identity \(\cos{x}=\cos{(180-x)}\) to get:

L.H.S.= \(\frac{\cos1^{\circ}-\frac{\cos89^{\circ}-\cos89^{\circ}}{2}}{\sin1^{\circ}}\)

L.H.S.= \(\frac{\cos1^{\circ}}{\sin1^{\circ}}=\cot1^{\circ}\) R.H.S.= \(\cot1^{\circ}\)

\(L.H.S.=R.H.S.\)

If \(\frac{2\cdot\sin2^{\circ}+4\cdot\sin4^{\circ}+\cdots+178\cdot\sin178^{\circ}+180\cdot\sin180^{\circ}}{90} = \cot 1^{\circ}\), that means the average of \(n \sin n^\circ\)(\(n = 2, 4, 6, \ldots, 180\)) is \(\cot 1^\circ\).

Therefore, the average of \(n \sin n^\circ\)(\(n = 2, 4, 6, \ldots, 180\))is \(\cot 1^\circ\).

 Jul 17, 2019

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