cos 66^{circ} + sin 84^{circ} = | vedclass

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(D) We know that $\sin 	heta = \cos(90^{\circ} - 	heta)$.Thus,$\sin 84^{\circ} = \cos(90^{\circ} - 84^{\circ}) = \cos 6^{\circ}$.Now,$\cos 66^{\circ

Vedclass · Accepted Answer

$\frac{1}{4} \sqrt{3}(\sqrt{5} + 1)$

Vedclass · Answer

(D) We know that $\sin 	heta = \cos(90^{\circ} - 	heta)$.Thus,$\sin 84^{\circ} = \cos(90^{\circ} - 84^{\circ}) = \cos 6^{\circ}$.Now,$\cos 66^{\circ} + \sin 84^{\circ} = \cos 66^{\circ} + \cos 6^{\circ}$.Using the formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$,we get:$= 2 \cos \frac{66^{\circ} + 6^{\circ}}{2} \cos \frac{66^{\circ} - 6^{\circ}}{2}$$= 2 \cos 36^{\circ} \cos 30^{\circ}$.Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$,we have:$= 2 	imes \left(\frac{\sqrt{5} + 1}{4}ight) 	imes \left(\frac{\sqrt{3}}{2}ight)$$= \frac{\sqrt{3}(\sqrt{5} + 1)}{4}$.Hence,option $D$ is correct.

$\cos 66^{\circ} + \sin 84^{\circ} = $

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