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July 2022 1 12 Report

Найти sin a, если cos a=12/13, 3п/2<a<2п

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RaiskaOmaMa Ответ:1)sina=-√1-144/169=-√25/169=-5/13 26sina=26*(-5/13)=-10 2)cos²a=1:(1+tg²a)=1:(1+2)=1/3 sin²a=1-cos²a=1-1/3=2/3 6sin²a=6*2/3=4 3)8tg(180-45)cos(360-60)=-8tg45cos60=-8*1*1/2=-4 4)12cos60=12*1/2=6

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