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 \(\cot 1^\circ\).
