已知i1=2sin(wt+30 )A,i2= 4sin(wt-45 )A,求i=i1+i2的最大值,要详细过程啊!~~急!

i=i1+i2=2sin(wt+30º)+4sin(wt-45º)
=2(sinwtcos30º+sin30coswt)+4(sinwtcos45º-sin45ºcoswt)
=√3sinwt+coswt+2√2sinwt-2√2coswt
=(√3+2√2)sinwt+(1-2√2)coswt
=√[(√3+2√2)²+(1-2√2)²]sin(wt+φ) （ 辅助角公式）
=√(3+8+4√6+1+8-4√2)sin(wt+φ)
=√(20+4√6-4√2)sin(wt+φ)
≤√(20+4√6-4√2)
所以i=i1+i2的最大值为√(20+4√6-4√2)。

i=i1+i2=2sin(wt+30º)+4sin(wt-45º)=2(sinwtcos30º+sin30coswt)+4(sinwtcos45º-sin45ºcoswt)=√3sinwt+coswt+2√2sinwt-2√2coswt=(√3+2√2)sinwt+(1-2√2)coswt=√[(√3+2√2)²+(1-...