Circuit Theory/Phasors/proof6

$g(t)=G_{m}sin(\omega t+\varphi )$

$g(t)=G_{m}cos(\omega t+\varphi -\frac{\pi }{2})$

$g(t)=G_m\mathrm{Re}(e^{j(\omega t+\varphi -\frac{\pi }{2})})$

$g(t)=G_m\mathrm{Re}(e^{j*(\varphi -\frac{\pi }{2})}{e}^{j\omega t})$

$g(t)=\mathrm{Re}(G_m{e}^{j*(\varphi -\frac{\pi }{2})}{e}^{j\omega t})$

$g(t)=\mathrm{Re}(𝔾e^{j\omega t})$

$𝔾=G_m{e}^{j*(\varphi -\frac{\pi }{2})}=G_m(cos(\varphi -\frac{\pi }{2})+j*sin(\varphi -\frac{\pi }{2}))$

$=G_{m}cos(\varphi -\frac{\pi }{2})+j{G}_{m}sin(\varphi -\frac{\pi }{2})$

$=G_m(cos(\varphi )cos(\frac{\pi }{2})+sin(\varphi )sin(\frac{\pi }{2}))+j{G}_m(sin(\varphi )cos(\frac{\pi }{2})-cos(\varphi )sin(\frac{\pi }{2}))$

$=G_{m}sin(\varphi )-j{G}_{m}cos(\varphi )$