Dado $\left|e^{i\frac{\phi}{2}}\left(e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}}\right)\right|$ demostrar que es igual a $2|\sin{\frac{\phi}{2}}|$

Question

Dado $\left|e^{i\frac{\phi}{2}}\left(e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}}\right)\right|$ Quiero demostrar que es igual a $2\left|\sin{\frac{\phi}{2}}\right|.$

Mi trabajo.

$|e^{i\frac{\phi}{2}}(e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}})|=|(\cos\frac{\phi}{2}+i\sin{\frac{\phi}{2}})(2i\sin{\frac{\phi}{2}})|=|\sin\phi i - 2\sin^2{\frac{\phi}{2}}|$ .

Cómo a partir de aquí pasar a demostrar que es igual a $2|\sin{\frac{\phi}{2}}|?$

Answer 1

$|e^{i\frac{\phi}{2}}(e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}})|=|e^{i\frac{\phi}{2}}|\times|e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}}|$ .

$|e^{i\frac{\phi}{2}}|=1$ .

Utilizar la identidad $e^{i\phi}-e^{-i\phi}=2i\sin\phi$ .

$|e^{i\frac{\phi}{2}}-e^{-i\frac{\phi}{2}}|=|2i\sin\frac{\phi}{2}|=2|\sin\frac{\phi}{2}|$ .

Answer 2

Continuación de $|i\sin\phi - 2\sin^2{\frac{\phi}{2}}|$ tenemos $$\sqrt{\sin^2\phi + 4\sin^4\frac{\phi}{2}} = \sqrt{4\sin^2\frac{\phi}{2}\cos^2\frac{\phi}{2} + 4\sin^4\frac{\phi}{2}} = \sqrt{4\sin^2\frac{\phi}{2}(\cos^2\frac{\phi}{2} + \sin^2\frac{\phi}{2}}) = \sqrt{4\sin^2\frac{\phi}{2}} = 2|\sin \frac{\phi}{2}|$$