Evaluate: [sqrt{2}(cos 56^{circ} 15^{prime} + | vedclass

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(D) Using De Moivre's Theorem,$[r(\cos 	heta + i \sin 	heta)]^n = r^n(\cos n	heta + i \sin n	heta)$.Given expression: $[\sqrt{2}(\cos 56^{\circ} 1

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$16i$

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(D) Using De Moivre's Theorem,$[r(\cos 	heta + i \sin 	heta)]^n = r^n(\cos n	heta + i \sin n	heta)$.Given expression: $[\sqrt{2}(\cos 56^{\circ} 15^{\prime} + i \sin 56^{\circ} 15^{\prime})]^8$.Here,$r = \sqrt{2}$,$	heta = 56^{\circ} 15^{\prime} = 56.25^{\circ}$,and $n = 8$.Applying the theorem:$= (\sqrt{2})^8 [\cos(8 	imes 56.25^{\circ}) + i \sin(8 	imes 56.25^{\circ})]$.$= 2^4 [\cos(450^{\circ}) + i \sin(450^{\circ})]$.Since $450^{\circ} = 360^{\circ} + 90^{\circ}$,we have $\cos(450^{\circ}) = \cos(90^{\circ}) = 0$ and $\sin(450^{\circ}) = \sin(90^{\circ}) = 1$.$= 16(0 + i(1)) = 16i$.

Evaluate: $[\sqrt{2}(\cos 56^{\circ} 15^{\prime} + i \sin 56^{\circ} 15^{\prime})]^8$

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