MHB How to Evaluate Trigonometric Cosine Sums Manually?

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Evaluate $\cos 5^{\circ}+\cos 77^{\circ}+\cos 149^{\circ}+\cos 221^{\circ}+\cos 293^{\circ}$ without the help of calculator.

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Congratulations to kaliprasad for his correct solutions::)

Spoiler

Using $\cos\,A + \cos\,B= 2\cos\dfrac{A+B}{2} \cos\dfrac{A-B}{2}\dots(1)$, we have

$\cos\,293^\circ + \cos\,77^\circ = 2 \cos\,185^\circ \cos\,108^\circ = 2 (- \cos\,5^\circ)(-\cos \,72^\circ)$

or $\cos\,293^\circ + \cos\,77^\circ = 2 \cos\,5^\circ \cos \,72^\circ\dots(2) $

further from (1)

$\cos\,221^\circ + \cos\,149^\circ = 2 \cos\,185^\circ \cos\,36^\circ = 2 (- \cos\,5^\circ)(\cos \,36^\circ)$

or $\cos\,221^\circ + \cos\,149^\circ = - 2 \cos\,5^\circ\cos \,36^\circ \dots(3)$

from (2) and (3) and adding $\cos\,5^\circ$

we get
$ \cos\,5^\circ + \cos\,77^\circ+ \cos\,149^\circ+ \cos\,221^\circ+ \cos\,293^\circ$
= $ \cos\,5^\circ + 2 \cos\,5^\circ \cos\,72^\circ -2 \cos\,5^\circ \cos\,36^\circ$
= $ \cos\,5^\circ ( 1 + 2 (\cos\,72^\circ - \cos\,36^\circ))$
= $ \cos\,5^\circ ( 1 - 2 * 2 (\sin \,54^\circ \sin \,18^\circ))$ using $\cos\,A - \cos\,B= 2\sin \dfrac{A+B}{2} \sin \dfrac{A-B}{2}$
= $\cos \,5^\circ( 1 - 4\dfrac{ \sin\, 2* 54^\circ}{2 * \cos\,54^\circ}\dfrac{ \sin\, 2* 18^\circ}{2 * \cos\,18^\circ})$
= $\cos \,5^\circ( 1 - \dfrac{ \sin\, 108 ^\circ}{\cos\,54^\circ}\dfrac{ \sin\, 36^\circ}{\cos\,18^\circ})$
= $\cos \,5^\circ( 1 - \dfrac{ \sin\, 72 ^\circ}{\sin\,36^\circ}\dfrac{ \sin\, 36^\circ}{\sin \,72^\circ})$
= $\cos \,5^\circ( 1 - 1)$
= 0