i1=100根号2sin(314t+45°),i2=60根号2sin(314t-30°),求i=i1+i2i2=60根号2sin(_数学解答

i1=100根号2sin(314t+45°),i2=60根号2sin(314t-30°),求i=i1+i2
i2=60根号2sin(314t-30°),

人气：290 ℃　时间：2019-10-19 17:00:56

解答

i=i1+i2
=100根号2sin(314t+45°)+60根号2sin(314t-30°)
=100根号2(sin314tcos45°+cos314tsin45°)+60根号2(sin314tcos30°-cos314tsin30°)
=100根号2(2分之根号2sin314t+2分之根号2cos314t)+60根号2(2分之根号3sin314t-2分之1cos314t)
=100分之1sin314t+100分之1cos314t+120分之根号6sin314t-120分之根号2cos314t
=600分之6sin314t+600分之6cos314t+600分之5根号6sin314t-600分之5根号2cos314t
=[600分之(6+5根号6)]sin314t+[600分之(6-5根号2)]cos314t