The value of sin 47^circ + sin 61^circ - sin | vedclass

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Question

(D) Given expression: $\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ$ Group the terms: $(\sin 61^\circ + \sin 47^\circ) - (\sin 25^\ci

Vedclass · Accepted Answer

$\cos 7^\circ $

Vedclass · Answer

(D) Given expression: $\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ$ Group the terms: $(\sin 61^\circ + \sin 47^\circ) - (\sin 25^\circ + \sin 11^\circ)$ Using the formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$: $= 2 \sin \frac{61^\circ + 47^\circ}{2} \cos \frac{61^\circ - 47^\circ}{2} - 2 \sin \frac{25^\circ + 11^\circ}{2} \cos \frac{25^\circ - 11^\circ}{2}$ $= 2 \sin 54^\circ \cos 7^\circ - 2 \sin 18^\circ \cos 7^\circ$ $= 2 \cos 7^\circ (\sin 54^\circ - \sin 18^\circ)$ Using the formula $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$: $= 2 \cos 7^\circ 	imes 2 \cos \frac{54^\circ + 18^\circ}{2} \sin \frac{54^\circ - 18^\circ}{2}$ $= 4 \cos 7^\circ \cos 36^\circ \sin 18^\circ$ Substitute values $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$ and $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$: $= 4 \cos 7^\circ \left( \frac{\sqrt{5}+1}{4} ight) \left( \frac{\sqrt{5}-1}{4} ight)$ $= 4 \cos 7^\circ \left( \frac{5-1}{16} ight) = 4 \cos 7^\circ \left( \frac{4}{16} ight) = \cos 7^\circ$

The value of $\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = $

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