Question

найдите :
1) sin2α,cos2α,ctg2α, если tgα =3/4

Answer 1

tga=3/4
cos²a=1:(1+tg²a)=1:(1+9/16)=1:25/16=16/25
cosa=4/5
sina=√(1-cos²a)=√(1-16/25)=√9/25=3/5
sin2a=2sinacosa=2*4/5*3/5=24/25
cos2a=cos²a-sin²a=16/25-9/25=7/25
ctg2a=cos2a/sin2a=7/25:24/25=7/25*25/24=7/24