The amplitude of sin frac{pi}{5} + i(1 - cos | vedclass

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Question

(C) Let $z = \sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$.Using trigonometric identities $\sin 2	heta = 2\sin 	heta \cos 	heta$ and $1 - \cos 2\

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$\pi/10$

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(C) Let $z = \sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$.Using trigonometric identities $\sin 2	heta = 2\sin 	heta \cos 	heta$ and $1 - \cos 2	heta = 2\sin^2 	heta$,we have:$z = 2\sin \frac{\pi}{10} \cos \frac{\pi}{10} + i(2\sin^2 \frac{\pi}{10})$$z = 2\sin \frac{\pi}{10} (\cos \frac{\pi}{10} + i\sin \frac{\pi}{10})$Since $2\sin \frac{\pi}{10} > 0$,the argument (amplitude) $	heta$ is given by:$	an 	heta = \frac{\sin(\pi/10)}{\cos(\pi/10)} = 	an \frac{\pi}{10}$Therefore,$	heta = \frac{\pi}{10}$.

The amplitude of $\sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$ is:

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