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複數的應用 (請詳細解答)

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http://i707.photobucket.com/albums/ww74/stevieg90/01-109.jpg (請詳細解答) 更新: 係唔係應該R1(t)+R2(t)=Acos(wt+z)??

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Denote θ1=x, θ2=y, θ3=z R1(t)+R2(t)=Re[10exp(iwt+ix)+8exp(iwt+iy)] 10exp(iwt+ix)+8exp(iwt+iy) =exp(iwt)[10exp(ix)+8exp(iy)] =exp(iwt)[(10cosx+8cosy)+i(10sinx+8siny)] =exp(iwt)[P+iQ] where P=10cosx+8cosy, Q=10sinx+8siny =exp(iwt)*A*[cosz+isinz] where A^2=P^2+Q^2, cosz=P/A, sinz=Q/A, tanz=Q/P =A exp(iwt)*exp(iz)=Aexp[i(wt+z)] so, R1(t)+R2(t)=Acos(wt-z) For y=10cos(t+π/6)+8cos(t+π/3) P=10cos(π/6)+8cos(π/3)=5√3+4 Q=10sin(π/6)+8sin(π/3)=5+4√3 A^2=100+64+80√3= 164+80√3, A=2√(41+20√3) tanz=Q/P, then z=arctan[(5+4√3)/(5√3+4)] about 43.3 degree. y=10cos(t+π/6)+8cos(t+π/3)=Acos(t-z) its graph is a cosinewave with amplitude A, period 2π and shifting to the right z units. 2010-06-24 13:30:47 補充： Thks! R1(t)+R2(t)=Acos(wt+z) y=Acos(t+z) shifting to the left z units.

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