已知u1=10sin(314t+π/2)V
已知u1=10sin(314t+π/2)V，u2=10sin(314t+π/3)V，试分别用相量求：
1）u1 + u2
2） u1－u2
u₁+u₂
=10[sin(314t+π/2)+sin(314t+π/3)]
=10*2sin[314t+(π/2+π/3)/2]cos[π/2-π/3)/2]
=20cos(π/12)sin(314t+5π/12)
=5(√6+√2)sin(314t+5π/12)
u₁-u₂
=10[sin(314t+π/2)-sin(314t+π/3)]
=10*2cos[314t+(π/2+π/3)/2]sin[(π/2-π/3)/2]
=20sin(π/12)cos(314t+5π/12)
=5(√6-√2)sin(314t+11π/12)
希望以下回答对您有帮助：u₁+u₂
=10[sin(314t+π/2)+sin(314t+π/3)]
=10*2sin[314t+(π/2+π/3)/2]cos[π/2-π/3)/2]
=20cos(π/12)sin(314t+5π/12)
=5(√6+√2)sin(314t+5π/12)u₁-u₂
=10[sin(314t+π/2)-sin(314t+π/3)]
=10*2cos[314t+(π/2+π/3)/2]sin[(π/2-π/3)/2]
=20sin(π/12)cos(314t+5π/12)
=5(√6-√2)sin(314t+11π/12)
表示向量的上箭头打不出来，以下用右箭头代替。
1、u = u1 + u2；
u1→ （ π/2，10 )，u2→ （ π/3，5√3 )；
u→ = u1→ + u2→ = ( π/2 + π/3，10 + 5√3 )
Asin(5π/6) = 10 + 5√3；
A = ( 10 + 5√3 )/sin(5π/6) = ( 10 + 5√3 )/sin(π/6) = 20 + 10√3
u1 + u2 = 10( 2 + √3 )sin( 314t + 5π/6 )；
2、u = u1 - u2；
u→ = u1→ - u2→ = ( π/2 - π/3，10 - 5√3 )；
Asin(π/6) = 10 - 5√3；
A = ( 10 - 5√3 )/sin(π/6) = 20 - 10√3
u1 - u2 = 10( 2 - √3 )sin( 314t + π/6 )；
上一篇：忠县白公祠是原址吗