left[frac{1+cos left(frac{pi}{12}right)+i sin | vedclass

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Question

(C) આપેલ પદને ધ્યાનમાં લો: $\left[\frac{\left(1+\cos \frac{\pi}{12}\right)+i \sin \frac{\pi}{12}}{\left(1+\cos \frac{\pi}{12}\right)-i \sin \frac{\pi}

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) આપેલ પદને ધ્યાનમાં લો: $\left[\frac{\left(1+\cos \frac{\pi}{12}ight)+i \sin \frac{\pi}{12}}{\left(1+\cos \frac{\pi}{12}ight)-i \sin \frac{\pi}{12}}ight]^{72}$અડધા ખૂણાના નિત્યસમ $1+\cos 	heta = 2 \cos^2 \frac{	heta}{2}$ અને $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ નો ઉપયોગ કરતા:$\left[\frac{2 \cos^2 \frac{\pi}{24} + i 2 \sin \frac{\pi}{24} \cos \frac{\pi}{24}}{2 \cos^2 \frac{\pi}{24} - i 2 \sin \frac{\pi}{24} \cos \frac{\pi}{24}}ight]^{72}$અંશ અને છેદમાંથી $2 \cos \frac{\pi}{24}$ સામાન્ય લેતા:$\left[\frac{2 \cos \frac{\pi}{24} (\cos \frac{\pi}{24} + i \sin \frac{\pi}{24})}{2 \cos \frac{\pi}{24} (\cos \frac{\pi}{24} - i \sin \frac{\pi}{24})}ight]^{72} = \left(\frac{\cos \frac{\pi}{24} + i \sin \frac{\pi}{24}}{\cos \frac{\pi}{24} - i \sin \frac{\pi}{24}}ight)^{72}$ગુણધર્મ $\frac{e^{i	heta}}{e^{-i	heta}} = e^{i2	heta}$ નો ઉપયોગ કરતા:$\left(e^{i \frac{\pi}{24}} / e^{-i \frac{\pi}{24}}ight)^{72} = (e^{i \frac{2\pi}{24}})^{72} = (e^{i \frac{\pi}{12}})^{72}$$= e^{i \frac{72\pi}{12}} = e^{i 6\pi}$ઓઈલરના સૂત્ર મુજબ,$e^{i 6\pi} = \cos(6\pi) + i \sin(6\pi) = 1 + i(0) = 1$.

$\left[\frac{1+\cos \left(\frac{\pi}{12}\right)+i \sin \left(\frac{\pi}{12}\right)}{1+\cos \left(\frac{\pi}{12}\right)-i \sin \left(\frac{\pi}{12}\right)}\right]^{72}=$

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