Question

cosa=3/5 cos2a-? sin2a-? tg2a-? a пренадлежит 4 четверти

Answer 1

cosa=3/5, a∈(3pi/2;2pi)
sina<0
cos^2a=9/25
cos^2a=1-sin^2a
sin^2a=1-cos^2a
sin^2a=1-9/25
sin^2a=16/25
sina= -4/5
sin2a=2sinacosa=2*(-4/5)*(3/5)= -24/25
tg2a=sin2a/cos2a= -24/25/(cos^2a-sin^2a)= -24/25/(9/25-16/25)= -24/-7=24/7

Answer 2

$cos alpha =frac{3}{5}; alpha in IV\sin alpha =-sqrt{1-cos^2 alpha }=-sqrt{frac{16}{25}}=-frac{4}{5}\cos2 alpha =cos^2 alpha -sin^2 alpha =frac{9}{25}-frac{16}{25}=-frac{7}{25}\sin2 alpha =2sin alpha cos alpha =-2*frac{4}{5}*frac{3}{5}=-frac{24}{25}\tg2 alpha =frac{sin2 alpha }{cos2 alpha }=frac{24}{25}*frac{25}{7}=frac{24}{7}=3frac{3}{7}$