Пожалуйста решите :
1.Найти 39 cos2α, если sin2 α=17/39
(sin2 α)^2=(17/39)^2
(cos2α)^2=1-(sin2 α)^2=1-(17/39)^2
cos2α=(√(39^2-17^2))/39
39cos2α=39*(√(39^2-17^2))/39=√(39^2-17^2)=√(1521-289)=√1232=4√77
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(sin2 α)^2=(17/39)^2
(cos2α)^2=1-(sin2 α)^2=1-(17/39)^2
cos2α=(√(39^2-17^2))/39
39cos2α=39*(√(39^2-17^2))/39=√(39^2-17^2)=√(1521-289)=√1232=4√77