The real part of (cos 4 + i sin 4 + 1)^{2020} | vedclass

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(B) Given expression: $(\cos 4 + i \sin 4 + 1)^{2020}$.Using the identities $1 + \cos 2A = 2 \cos^2 A$ and $\sin 2A = 2 \sin A \cos A$,we have:$(\cos

Vedclass · Accepted Answer

$2^{2020} \cos^{2020} 2 \cos 4040$

Vedclass · Answer

(B) Given expression: $(\cos 4 + i \sin 4 + 1)^{2020}$.Using the identities $1 + \cos 2A = 2 \cos^2 A$ and $\sin 2A = 2 \sin A \cos A$,we have:$(\cos 4 + 1) + i \sin 4 = 2 \cos^2 2 + 2i \sin 2 \cos 2$.Factoring out $2 \cos 2$,we get:$2 \cos 2 (\cos 2 + i \sin 2)$.Raising this to the power of $2020$:$[2 \cos 2 (\cos 2 + i \sin 2)]^{2020} = 2^{2020} \cos^{2020} 2 (\cos 2 + i \sin 2)^{2020}$.Applying De Moivre's Theorem,$(\cos 	heta + i \sin 	heta)^n = \cos(n	heta) + i \sin(n	heta)$:$2^{2020} \cos^{2020} 2 (\cos(2020 	imes 2) + i \sin(2020 	imes 2)) = 2^{2020} \cos^{2020} 2 (\cos 4040 + i \sin 4040)$.The real part is $2^{2020} \cos^{2020} 2 \cos 4040$.

The real part of $(\cos 4 + i \sin 4 + 1)^{2020}$ is $.........$

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