Suppose that v1(t)=100cos(ωt)and v2(t)=100sin(ωt). Use phasors to reduce the sum vs(t)=v1(t)+v2(t) to a single term of the form Vmcos(ωt+θ). Enter your answers numerically separated by a comma. Express θ in degrees using three significant figures

Answer:[tex]vs(t)=152.08cos(\omega t -45)=152.08\angle-45[/tex]Step-by-step explanation:A tension expressed in form:[tex]V(t)=Vm*cos(\omega t+\phi)[/tex]It can be converted directly into a fasor:[tex]V(t)=Vm\angle \phi[/tex]If the function is a positive sine we subtract 90 °[tex]V(t)=Vm*sin(\omega t+\phi)=Vm*cos(\omega t+\phi-90)[/tex]So:[tex]v1(t)=100cos(\omega t)= 100\angle0\\v2(t)=100sin(\omega t)=100cos(\omega t-\ph90)=100\angle -90[/tex][tex]vs(t)=v1(t)+v2(t)=100\angle 0 +100\angle-90=152.0811931\angle-45\approx152.08\angle-45[/tex]Therefore:[tex]vs(t)=152.08\angle-45=152.08cos(\omega t -45)[/tex]