For n in Z^{+},(1+sin theta+i cos theta)^n+(1 | vedclass

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Question

(C) Let $z = 1+\sin 	heta+i \cos 	heta$. We can write this as $z = 1+\cos(\frac{\pi}{2}-	heta) + i\sin(\frac{\pi}{2}-	heta)$.Using the identities

Vedclass · Accepted Answer

$2^{n+1} \cdot \cos ^n\left(\frac{\pi}{4}-\frac{	heta}{2}ight) \cos \left(\frac{n \pi}{4}-\frac{n 	heta}{2}ight)$

Vedclass · Answer

(C) Let $z = 1+\sin 	heta+i \cos 	heta$. We can write this as $z = 1+\cos(\frac{\pi}{2}-	heta) + i\sin(\frac{\pi}{2}-	heta)$.Using the identities $1+\cos(2A) = 2\cos^2(A)$ and $\sin(2A) = 2\sin(A)\cos(A)$,where $A = \frac{\pi}{4}-\frac{	heta}{2}$,we get:$z = 2\cos^2(\frac{\pi}{4}-\frac{	heta}{2}) + 2i\sin(\frac{\pi}{4}-\frac{	heta}{2})\cos(\frac{\pi}{4}-\frac{	heta}{2})$$z = 2\cos(\frac{\pi}{4}-\frac{	heta}{2}) [\cos(\frac{\pi}{4}-\frac{	heta}{2}) + i\sin(\frac{\pi}{4}-\frac{	heta}{2})]$Then $z^n = 2^n \cos^n(\frac{\pi}{4}-\frac{	heta}{2}) [\cos(\frac{n\pi}{4}-\frac{n	heta}{2}) + i\sin(\frac{n\pi}{4}-\frac{n	heta}{2})]$.Similarly,the conjugate term is $\bar{z}^n = 2^n \cos^n(\frac{\pi}{4}-\frac{	heta}{2}) [\cos(\frac{n\pi}{4}-\frac{n	heta}{2}) - i\sin(\frac{n\pi}{4}-\frac{n	heta}{2})]$.Adding these two expressions:$z^n + \bar{z}^n = 2^n \cos^n(\frac{\pi}{4}-\frac{	heta}{2}) [2\cos(\frac{n\pi}{4}-\frac{n	heta}{2})]$$= 2^{n+1} \cos^n(\frac{\pi}{4}-\frac{	heta}{2}) \cos(\frac{n\pi}{4}-\frac{n	heta}{2})$.

For $n \in Z^{+}$,$(1+\sin \theta+i \cos \theta)^n+(1+\sin \theta-i \cos \theta)^n=$

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