Evaluate: (cos 252^{circ} - sin 126^{circ})(c | vedclass

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(B) Let $E = (\cos^2 252^{\circ} - \sin^2 126^{\circ})(\sin^2 126^{\circ} + \sin^2 186^{\circ} + \sin^2 66^{\circ})$.Using $\cos^2 	heta = \frac{1+\c

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$\frac{-3 \sqrt{5}}{8}$

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(B) Let $E = (\cos^2 252^{\circ} - \sin^2 126^{\circ})(\sin^2 126^{\circ} + \sin^2 186^{\circ} + \sin^2 66^{\circ})$.Using $\cos^2 	heta = \frac{1+\cos 2	heta}{2}$ and $\sin^2 	heta = \frac{1-\cos 2	heta}{2}$:$\cos^2 252^{\circ} - \sin^2 126^{\circ} = \frac{1+\cos 504^{\circ}}{2} - \frac{1-\cos 252^{\circ}}{2} = \frac{\cos 504^{\circ} + \cos 252^{\circ}}{2} = \frac{2 \cos 378^{\circ} \cos 126^{\circ}}{2} = \cos 18^{\circ} \cos 126^{\circ} = \cos 18^{\circ} (-\sin 36^{\circ})$.Now,$\sin^2 126^{\circ} + \sin^2 186^{\circ} + \sin^2 66^{\circ} = \frac{1-\cos 252^{\circ}}{2} + \frac{1-\cos 372^{\circ}}{2} + \frac{1-\cos 132^{\circ}}{2} = \frac{3 - (\cos 252^{\circ} + \cos 12^{\circ} + \cos 132^{\circ})}{2}$.Using $\cos 252^{\circ} + \cos 132^{\circ} = 2 \cos 192^{\circ} \cos 60^{\circ} = \cos 192^{\circ} = -\cos 12^{\circ}$,the sum becomes $\frac{3 - (-\cos 12^{\circ} + \cos 12^{\circ})}{2} = \frac{3}{2}$.Thus,$E = \cos 18^{\circ} (-\sin 36^{\circ}) 	imes \frac{3}{2} = -\frac{3}{2} \sin 36^{\circ} \cos 18^{\circ} = -\frac{3}{2} 	imes \frac{2 \sin 18^{\circ} \cos 18^{\circ} \cos 18^{\circ}}{\sin 18^{\circ}}$ (or simply use $\sin 36^{\circ} = 2 \sin 18^{\circ} \cos 18^{\circ}$).$E = -\frac{3}{2} (2 \sin 18^{\circ} \cos 18^{\circ}) \cos 18^{\circ} = -3 \sin 18^{\circ} \cos^2 18^{\circ} = -3 \sin 18^{\circ} (1 - \sin^2 18^{\circ})$.Using $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$,$E = -3 \left(\frac{\sqrt{5}-1}{4}ight) \left(1 - \frac{6-2\sqrt{5}}{16}ight) = -\frac{3(\sqrt{5}-1)}{4} 	imes \frac{10+2\sqrt{5}}{16} = -\frac{3(10\sqrt{5} + 10 - 10 - 2\sqrt{5})}{64} = -\frac{3(8\sqrt{5})}{64} = -\frac{3\sqrt{5}}{8}$.

Evaluate: $(\cos 252^{\circ} - \sin 126^{\circ})(\cos 252^{\circ} + \sin 126^{\circ})(\sin^2 126^{\circ} + \sin^2 186^{\circ} + \sin^2 66^{\circ})$

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