Simplify $\cos 1^\circ + \cos 3^\circ + \cdots+ \cos 43^\circ$? I am currently working on a problem and reduced part of the equations down to
$$\cos(1^\circ)+\cos(3^\circ)+\cdots+\cos(39^\circ)+\cos(41^\circ)+\cos(43^\circ)$$
How can I calculate this easily using the product-to-sum formula for $\cos(x)+\cos(y)$?
 A: The trick is to multiply the whole sum by $2\sin 1^\circ$. Since:
$$ 2\sin 1^\circ \cos 1^{\circ} = \sin 2^\circ - \sin 0^\circ,$$
$$ 2\sin 1^\circ \cos 3^{\circ} = \sin 4^\circ - \sin 2^\circ,$$
$$\ldots $$
$$ 2\sin 1^\circ \cos 43^{\circ} = \sin 44^\circ - \sin 42^\circ,$$
by adding these identites we get that the original sum $S$, multiplied by $2\sin 1^\circ$, equals $\sin 44^\circ$.
This just gives $S=\color{red}{\frac{\sin 44^\circ}{2\sin 1^{\circ}}}.$

Footnote. Such sum must be greater than $\frac{44\sqrt{2}}{\pi}$, but not greater than $\frac{44\sqrt{2}}{\pi}+\frac{\pi}{22\sqrt{2}}$, by a Riemann sum $+$ concavity argument.
A: let $S = \cos(1^\circ)+\cos(3^\circ)+.....+\cos(39^\circ)+\cos(41^\circ)+\cos(43^\circ).$ then 
\begin{align} 2S\sin 1^\circ &= 2\cos 1^\circ \sin 1^\circ + 2 \cos 3^\circ \sin 1^\circ+\cdots + 2 \cos 43^\circ \sin 1^\circ \\
& = (\sin 2^\circ - \sin 0^\circ)+(\sin 4^\circ - \sin 2^\circ ) + \cdots 
+(\sin 44^\circ- \sin 42^\circ))\\
& = \sin44^\circ 
\end{align}
therefore $$ S = \dfrac{\sin 44^\circ}{ 2\sin 1^\circ} $$
