MCQ
$\sin {47^o} + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = $ . . . .
- A$\sin 36^\circ $
- B$\cos 36^\circ $
- C$\sin 7^\circ $
- ✓$\cos 7^\circ $
$= 2 sin 54^\circ cos 7^\circ - 2 sin 18^\circ cos 7^\circ$
$ = \,\,2\,\,\cos \,\,{7^o}\,(\sin \,\,{54^o} - \sin \,\,{18^o})$
$ = \,\,2\,\,\cos \,\,{7^o}\,\,.\,\,2\,\,\cos \,\,{36^o}\,\,.\,\,\sin \,\,{18^o}$
$ = \,\,4.\,\cos \,\,{7^o}.\,\frac{{\sqrt 5 + 1}}{4}.\frac{{\sqrt 5 - 1}}{4} = \cos \,\,{7^{o.}}$.
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