Davis

Thanks!

Just wondering, did you use a computer to find this?

Thanks!

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 $\cot 1^\circ$.

OK, I wont delete my stuff next time

Thanks!

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

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

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