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$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $
$2\,\cos {28^o}\,\cos \,{29^o}\,\cos \,{33^o}$
$4\,\cos {28^o}\,\cos \,{29^o}\,\cos \,{33^o}$
$4\,\cos {28^o}\,\cos \,{29^o}\,\sin {33^o}$
$2\,\cos {28^o}\,\cos \,{29^o}\,\sin \,{33^o}$
Solution
(c) $1 + \cos 56^\circ + \cos 58^\circ – \cos 66^\circ $
$ = 2{\cos ^2}28^\circ + 2\sin 62^\circ .\sin 4^\circ $
$ = 2{\cos ^2}28^\circ + 2\cos 28^\circ .\sin 4^\circ $
$ = 2\cos 28^\circ (\cos 28^\circ + \cos 86^\circ )$
$ = 2\cos 28^\circ .2\cos 57^\circ \cos 29^\circ $
$ = 4\cos 28^\circ \cos 29^\circ \sin 33^\circ $.
Aliter : Apply the conditional identity
$\cos A + \cos B – \cos C = – 1 + 4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$ $[\, \because 56^\circ + 58^\circ + 66^\circ = 180^\circ ]$
We get the value of required expression equal to $4\cos 28^\circ \cos 29^\circ \sin 33^\circ $.
