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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 $\cos(x)+\cos(y)$?

Leucippus
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Mathy Person
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  • It is an abstract duplicate, but I do not agree about closing it. Since it is an abstract duplicate, not an exact or almost exact one. – Jack D'Aurizio Jan 12 '15 at 21:28

2 Answers2

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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.

Jack D'Aurizio
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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 $$ S = \dfrac{\sin 44^\circ}{ 2\sin 1^\circ} $$

abel
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