Question

ПУсть sina=5/13, a∈ II четверти найти sin2a, cos2a

Answer 1

$sin2a=2siaacosa,co2a=cos ^{2}a-sin ^{2}a cosa= - \sqrt{1- \frac{25}{169} } =- \sqrt{ \frac{144}{169}$$cosa=- \frac{12}{13} , sin2a=2 \frac{5}{13} ( \frac{ -12}{13} )=- \frac{120}{169}$$cos2a=cos ^{2} a-sin ^{2} a, cos2a= \frac{144}{169}- \frac{25}{169}= \frac{119}{169}$