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olgapatayka59
@olgapatayka59
July 2022
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Дано:cosα=5/13 0 меньше α меньше π/2. Найти sin 2α?
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Gerren
Sin2a=2sinacosa
sina=√(1-cosa^2)
sina=√(1-(5/13)^2)=√(1-25/169)=√144/169=12/13 тк 0<a<pi/2 - 1 четверть sina>0
sin2a=2*5/13*12/13=120/169
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Answers & Comments
sina=√(1-cosa^2)
sina=√(1-(5/13)^2)=√(1-25/169)=√144/169=12/13 тк 0<a<pi/2 - 1 четверть sina>0
sin2a=2*5/13*12/13=120/169