{left( frac{1 + cos phi + isin phi }{1 + cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given expression: $L.H.S. = {\left[ \frac{1 + \cos \phi + i\sin \phi }{1 + \cos \phi - i\sin \phi } \right]^n}$Using half-angle identities $1 + \c

Vedclass · Accepted Answer

$\cos n\phi + i\sin n\phi $

Vedclass · Answer

(B) Given expression: $L.H.S. = {\left[ \frac{1 + \cos \phi + i\sin \phi }{1 + \cos \phi - i\sin \phi } ight]^n}$Using half-angle identities $1 + \cos \phi = 2\cos^2(\phi/2)$ and $\sin \phi = 2\sin(\phi/2)\cos(\phi/2)$:$L.H.S. = {\left[ \frac{2\cos^2(\phi/2) + 2i\sin(\phi/2)\cos(\phi/2)}{2\cos^2(\phi/2) - 2i\sin(\phi/2)\cos(\phi/2)} ight]^n}$$= {\left[ \frac{2\cos(\phi/2)(\cos(\phi/2) + i\sin(\phi/2))}{2\cos(\phi/2)(\cos(\phi/2) - i\sin(\phi/2))} ight]^n}$$= {\left[ \frac{\cos(\phi/2) + i\sin(\phi/2)}{\cos(\phi/2) - i\sin(\phi/2)} ight]^n}$Using Euler's formula $e^{i	heta} = \cos 	heta + i\sin 	heta$:$= {\left[ \frac{e^{i\phi/2}}{e^{-i\phi/2}} ight]^n} = {(e^{i\phi/2 + i\phi/2})^n} = {(e^{i\phi})^n}$$= e^{in\phi} = \cos n\phi + i\sin n\phi$.

${\left( \frac{1 + \cos \phi + i\sin \phi }{1 + \cos \phi - i\sin \phi } \right)^n} = $

Similar Questions

${\left[ {\frac{{1 + \cos (\pi /8) + i\sin (\pi /8)}}{{1 + \cos (\pi /8) - i\sin (\pi /8)}}} \right]^8}$ is equal to

If $n, K \in N$ such that $n \neq 3K$,then $(\sqrt{3}+i)^{2n} + (\sqrt{3}-i)^{2n} = $

If $1, z_1, z_2, \ldots, z_{n-1}$ are the $n$th roots of unity,then the value of $(1-z_1)(1-z_2) \ldots (1-z_{n-1})$ is equal to

If $\omega$ is a complex cube root of unity,then for any $n>1$,$\sum_{r=1}^{n-1} r(r+1-\omega)(r+1-\omega^2) =$

$(-\sqrt{3} + i)^{53}$,where $i^2 = -1$,is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module