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alligator02nd
@alligator02nd
August 2022
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Помогите, пожалуйста, очень нужно решить прямо сейчас. Хотя бы первые два выражения
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sedinalana
Пищу сразу решение
1
2*1/2*(cos(a+b-a+b)-cos(a+b+a-b))+cos2a=cos2b-cos2a-cos2a=cos2b
2
cos4a:(sin4a-sin2a/cos2a)=cos4a*cos2a/(sin4acos2a-sin2a)=
=cos4acos2a/(2sin2acos²2a-sin2a)=cos4acos2a/[sin2a(2cos²2a-1]=
=cos4acos2a/(sin2a*cos4a)=cos2a/sin2a=ctg2a
3
2sin2acos3a/2cos3a*cosa/sina - 1=sin2acosa/sina - 1=
=2sinacos²a/sina -1=2cos²a-1=cos2a
4
2sin2acosa/2cos2acosa *(1+cos4a)=sin2a/cos2a * 2cos²2a=
=2sin2acos2a=sin4a
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Answers & Comments
1
2*1/2*(cos(a+b-a+b)-cos(a+b+a-b))+cos2a=cos2b-cos2a-cos2a=cos2b
2
cos4a:(sin4a-sin2a/cos2a)=cos4a*cos2a/(sin4acos2a-sin2a)=
=cos4acos2a/(2sin2acos²2a-sin2a)=cos4acos2a/[sin2a(2cos²2a-1]=
=cos4acos2a/(sin2a*cos4a)=cos2a/sin2a=ctg2a
3
2sin2acos3a/2cos3a*cosa/sina - 1=sin2acosa/sina - 1=
=2sinacos²a/sina -1=2cos²a-1=cos2a
4
2sin2acosa/2cos2acosa *(1+cos4a)=sin2a/cos2a * 2cos²2a=
=2sin2acos2a=sin4a