Evaluate: sin 6^{circ} + sin 54^{circ} + sin | vedclass

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Question

(A) Given expression: $E = \sin 6^{\circ} + \sin 54^{\circ} + \sin 126^{\circ} + \cos 156^{\circ}$ Using the identity $\sin(180^{\circ} - 	heta) = \s

Vedclass · Accepted Answer

$\frac{\sqrt{5}+1}{4}$

Vedclass · Answer

(A) Given expression: $E = \sin 6^{\circ} + \sin 54^{\circ} + \sin 126^{\circ} + \cos 156^{\circ}$ Using the identity $\sin(180^{\circ} - 	heta) = \sin 	heta$ and $\cos(180^{\circ} - 	heta) = -\cos 	heta$: $\sin 126^{\circ} = \sin(180^{\circ} - 54^{\circ}) = \sin 54^{\circ}$ $\cos 156^{\circ} = \cos(180^{\circ} - 24^{\circ}) = -\cos 24^{\circ}$ So,$E = \sin 6^{\circ} + 2 \sin 54^{\circ} - \cos 24^{\circ}$ Using $\sin 54^{\circ} = \cos 36^{\circ}$ and $\cos 24^{\circ} = \sin 66^{\circ}$: $E = \sin 6^{\circ} - \sin 66^{\circ} + 2 \cos 36^{\circ}$ Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$: $E = 2 \cos 36^{\circ} \sin(-30^{\circ}) + 2 \cos 36^{\circ}$ $E = 2 \cos 36^{\circ} 	imes (-\frac{1}{2}) + 2 \cos 36^{\circ}$ $E = -\cos 36^{\circ} + 2 \cos 36^{\circ} = \cos 36^{\circ}$ Since $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$,the result is $\frac{\sqrt{5}+1}{4}$.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

Evaluate: $\sin 6^{\circ} + \sin 54^{\circ} + \sin 126^{\circ} + \cos 156^{\circ}$

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