Вычислите а) sin (arctg 2/3) б

Объяснение:

а)tg a = 2/3; tg^2 a = 4/9; 1 + tg^2 a = 1/cos^2 a

cos^2 a = 1 / (1 + tg^2 a) = 1 / (1 + 4/9) = 1 / (13/9) = 9/13

sin^2 a = 1 - cos^2 a = 1 - 9/13 = 4/13

sin a = √(4/13) = 2/√13 = 2√13/13

б)cos (2arccos 1/4)==2cos^2(arccos1/4)-1=2*1/16-1=1/8-1=-7/8