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

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(B) આપેલ પદાવલિ: $L.H.S. = {\left[ \frac{1 + \cos \phi + i\sin \phi }{1 + \cos \phi - i\sin \phi } \right]^n}$અડધા ખૂણાના નિત્યસમ $1 + \cos \phi = 2\c

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) આપેલ પદાવલિ: $L.H.S. = {\left[ \frac{1 + \cos \phi + i\sin \phi }{1 + \cos \phi - i\sin \phi } ight]^n}$અડધા ખૂણાના નિત્યસમ $1 + \cos \phi = 2\cos^2(\phi/2)$ અને $\sin \phi = 2\sin(\phi/2)\cos(\phi/2)$ નો ઉપયોગ કરતા:$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}$ઓઈલરના સૂત્ર $e^{i	heta} = \cos 	heta + i\sin 	heta$ નો ઉપયોગ કરતા:$= {\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} = $

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