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drwnd
@drwnd
July 2022
1
26
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помогите представить в стандартной тригонометрической форме:
[tex]1-cos(250)+isin(610)[/tex]
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Gerren
Z=|z|(cosФ+isinФ)
1=x+i*0
|z|=√(x^2+y^2)
|z|=1
arg(z)=Ф=arxtg(y/x)=0/1=0
1=cos0+isin0=cos0
cos250=cos(180+70)=-cos70
sin610=sin(360+180+70)=-sin70
cos0+cos70=-2(cos35)^2
Ф=arctg(-sin70/-2(cos35)^2)=arctg(sin70/2(cos35)^2)
|z|=√(-2(cos35)^2)^2+(sin70)^2)=√4(cos35)^4+(sin70)^2)
z=√4(cos35)^4+(sin70)^2)(-2(cos35)^2+isin70)
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Answers & Comments
1=x+i*0
|z|=√(x^2+y^2)
|z|=1
arg(z)=Ф=arxtg(y/x)=0/1=0
1=cos0+isin0=cos0
cos250=cos(180+70)=-cos70
sin610=sin(360+180+70)=-sin70
cos0+cos70=-2(cos35)^2
Ф=arctg(-sin70/-2(cos35)^2)=arctg(sin70/2(cos35)^2)
|z|=√(-2(cos35)^2)^2+(sin70)^2)=√4(cos35)^4+(sin70)^2)
z=√4(cos35)^4+(sin70)^2)(-2(cos35)^2+isin70)