If frac{pi}{2}

Question

(C) Let $z = (1 + \cos 2\alpha) + i \sin 2\alpha$.Using trigonometric identities,$1 + \cos 2\alpha = 2 \cos^2 \alpha$ and $\sin 2\alpha = 2 \sin \alph

Vedclass · Accepted Answer

$-2 \cos \alpha, \alpha - \pi$

Vedclass · Answer

(C) Let $z = (1 + \cos 2\alpha) + i \sin 2\alpha$.Using trigonometric identities,$1 + \cos 2\alpha = 2 \cos^2 \alpha$ and $\sin 2\alpha = 2 \sin \alpha \cos \alpha$.So,$z = 2 \cos^2 \alpha + i(2 \sin \alpha \cos \alpha) = 2 \cos \alpha (\cos \alpha + i \sin \alpha)$.Given $\frac{\pi}{2} $|z| = |2 \cos \alpha| = -2 \cos \alpha$.To express $z$ in polar form $r(\cos 	heta + i \sin 	heta)$ where $r > 0$,we write:$z = -2 \cos \alpha (-\cos \alpha - i \sin \alpha)$.Since $-\cos \alpha = \cos(\alpha - \pi)$ and $-\sin \alpha = \sin(\alpha - \pi)$,$z = -2 \cos \alpha (\cos(\alpha - \pi) + i \sin(\alpha - \pi))$.Therefore,the modulus is $-2 \cos \alpha$ and the argument is $\alpha - \pi$.

If $\frac{\pi}{2} < \alpha < \frac{3\pi}{2}$,then the modulus and argument of $(1 + \cos 2\alpha) + i \sin 2\alpha$ are respectively:

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