The value of left[ frac{1 - cos frac{pi}{10} | vedclass

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Question

(B) Let $	heta = \frac{\pi}{10}$. The expression is $\left[ \frac{1 - \cos 	heta + i\sin 	heta}{1 - \cos 	heta - i\sin 	heta} ight]^{10}$.Using

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$-1$

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(B) Let $	heta = \frac{\pi}{10}$. The expression is $\left[ \frac{1 - \cos 	heta + i\sin 	heta}{1 - \cos 	heta - i\sin 	heta} ight]^{10}$.Using the identities $1 - \cos 	heta = 2\sin^2 \frac{	heta}{2}$ and $\sin 	heta = 2\sin \frac{	heta}{2} \cos \frac{	heta}{2}$:Numerator $= 2\sin^2 \frac{	heta}{2} + i(2\sin \frac{	heta}{2} \cos \frac{	heta}{2}) = 2\sin \frac{	heta}{2} (\sin \frac{	heta}{2} + i\cos \frac{	heta}{2}) = 2\sin \frac{	heta}{2} [i(\cos \frac{	heta}{2} - i\sin \frac{	heta}{2})] = 2i\sin \frac{	heta}{2} e^{-i	heta/2}$.Denominator $= 2\sin^2 \frac{	heta}{2} - i(2\sin \frac{	heta}{2} \cos \frac{	heta}{2}) = 2\sin \frac{	heta}{2} (\sin \frac{	heta}{2} - i\cos \frac{	heta}{2}) = 2\sin \frac{	heta}{2} [-i(\cos \frac{	heta}{2} + i\sin \frac{	heta}{2})] = -2i\sin \frac{	heta}{2} e^{i	heta/2}$.Dividing the two: $\frac{2i\sin \frac{	heta}{2} e^{-i	heta/2}}{-2i\sin \frac{	heta}{2} e^{i	heta/2}} = -e^{-i	heta} = -(\cos 	heta - i\sin 	heta)$.Raising to the power of $10$: $(-1)^{10} (\cos 	heta - i\sin 	heta)^{10} = 1 \cdot (\cos 10	heta - i\sin 10	heta) = \cos \pi - i\sin \pi = -1 - 0 = -1$.

The value of $\left[ \frac{1 - \cos \frac{\pi}{10} + i\sin \frac{\pi}{10}}{1 - \cos \frac{\pi}{10} - i\sin \frac{\pi}{10}} \right]^{10} = $

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