The value of the expression frac{(sin 36^{cir | vedclass

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Question

(C) Let the expression be $E = \frac{(\sin 36^{\circ} + \cos 36^{\circ} - \sqrt{2} \sin 27^{\circ})^2}{2 \sin 54^{\circ}}$.Expanding the numerator: $(

Vedclass · Accepted Answer

$\cos 9^{\circ}$

Vedclass · Answer

(C) Let the expression be $E = \frac{(\sin 36^{\circ} + \cos 36^{\circ} - \sqrt{2} \sin 27^{\circ})^2}{2 \sin 54^{\circ}}$.Expanding the numerator: $(\sin 36^{\circ} + \cos 36^{\circ})^2 + 2 \sin^2 27^{\circ} - 2\sqrt{2} \sin 27^{\circ} (\sin 36^{\circ} + \cos 36^{\circ})$.Using $\sin 36^{\circ} + \cos 36^{\circ} = \sqrt{2} \sin(36^{\circ} + 45^{\circ}) = \sqrt{2} \sin 81^{\circ} = \sqrt{2} \cos 9^{\circ}$.Alternatively,simplify the expression as follows:$E = \frac{1 + \sin 72^{\circ} - 2\sqrt{2} \sin 27^{\circ} (\sin 36^{\circ} + \cos 36^{\circ})}{2 \cos 36^{\circ}}$.Using the identity $\sin 36^{\circ} + \cos 36^{\circ} = \sqrt{2} \cos 9^{\circ}$ and $\sin 27^{\circ} = \cos 63^{\circ}$,the expression simplifies to $\cos 18^{\circ}$.Since $\cos 18^{\circ} < \cos 9^{\circ}$,the value is less than $\cos 9^{\circ}$.

The value of the expression $\frac{(\sin 36^{\circ} + \cos 36^{\circ} - \sqrt{2} \sin 27^{\circ})^2}{2 \sin 54^{\circ}}$ is less than

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