Question

Sina=4/5;П/2<a<П найти cos2a,ctg2a

Answer 1

Cosa=-√1-(4/5)²=-√9/25=-3/5 т.к cos во втором квадранте
отрицателен
cos2a=2cos²a-1=2*(-3/5)²-1=2*9/25-1=18/25-1=
=-7/25
ctg2a=cos2a/sin2a=(-7/25)/2*sinacosa=(-7/25)/2*4/5*(-3/5)=(-7/25)/(-24/25)=7/24

Answer 2

Given that sin(a) = 4/5 and π/2 < a < π, we can use the Pythagorean identity to find cos(a): cos(a) = √(1 - sin^2(a)) cos(a) = √(1 - 16/25) cos(a) = √(9/25) cos(a) = 3/5

Now we can use the double angle formulas to find cos(2a) and ctg(2a): cos(2a) = 2cos^2(a) - 1 cos(2a) = 2(3/5)^2 - 1 cos(2a) = 2(9/25) - 1 cos(2a) = 18/25 - 25/25 cos(2a) = -7/25

ctg(2a) = (cos(2a))/sin(2a) ctg(2a) = (cos(2a))/(2sin(a)cos(a)) ctg(2a) = (-7/25)/(2(4/5)(3/5)) ctg(2a) = (-7/25)/(24/25) ctg(2a) = -7/24

Therefore, cos(2a) = -7/25 and ctg(2a) = -7/24.