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да, отношение sin7a к sina.
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Ответы и объяснения

Лучший Ответ!
2013-11-05T22:24:10+04:00
Sin7a/sina - -2(cos2a+cos4a+cos6a)-1=0;
sin(6a+a)/sina-2cos2--2cos4a-2cos6a-1=(sin6acosa+cos6asina/sina-2cos6a)-2cos2-2cos4a-1=(sin4acosa+cos4asina-2cos4asina/sina-2cos2a-1=sin2acosa+cos2asina-2cos2asina/sina-1=sina/sina-1;
1-1=0...
Тождество доказано!
2013-11-05T22:34:35+04:00
Sin(6a+a)/sina - 2cos2a-2cos4a-1=(sin6acosa+cos6asina/sina - 2cos6a) - 2cos2a-2cos4a-1=sin6acosa+cos6asina=2cos6asina/sina - 2cos2a-2cos4a-1=sin6acosa-cos6asina/sina - 2cos2a-2cos4a-1=(sin5a/sina - 2cos4a) - 2cos2a-1= ( sin(4a+a)/sina - 2cos4a) - 2cos2a-1=(sin4acosa+cos4asina/sina - 2cos4a) - 2cos2a-1= sin4acosa+cos4asina-2cos4asin/sina - 2cos2a-1=sin4acosa-cos4asina/sina-2cos2a-1=sin3a/sina-2cos2a-1=(sin(2a+a)/sina=2cos2a)-1=(sin2acosa+cos2asina/sina - 2cos2a) -1=sin2acosa+cos2asina-2cos2asina/sina-1=sin2acosa-cosa-cos2asina/sina-1=sina/sina-1=1-1=0.