+501 \(2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} = 90\cdot \cot 1^{\circ}\)
\(2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} = 90\cdot \frac{\cos 1}{\sin 1}\)
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
\(\cos1^{\circ}+sin1^{\circ}=\cos1^{\circ}\)
+118680 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?
+501 I'm trying to prove that the average of n \(n \sin n^\circ\)(\(n = 2, 4, 6, \ldots, 180\)) is \(\cot 1^\circ\)
+118680 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.
+501 ?
If I show that the equation simplifies to cos(1)=cos(1) or something, doesn't that work too?
+26393 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!
+26393 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} } \)
+26393 if line 3 is not wrong, why don't you get \(\cos 1^{\circ} = \cos 1^{\circ}\) ?
I got it out !
+501 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 \(\cot 1^\circ\).
