What did I do wrong?

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.

?

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

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.

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

?

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?

$2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} = 90\cdot \cot 1^{\circ}$

$\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} }$

Very nice, heureka !!!!

Thank You, CPhill !

Thanks!