(срочно) памагите решить !!! митематику

1) sin^2 (pi/2 - a) + sin^2 (pi + a) = cos^2 a + (-sin a)^2 = 12) sin 4a*cos a - cos 4a*sin a = sin (4a - a) = sin 3a3) cos a = 4/5; 0 < a < pi/2; значит, sin a > 0sin^2 a = 1 - cos^2 a = 1 - 16/25 = 9/25; sin a = 3/5sin 2a = 2sin a*cos a = 2*3/5*4/5 = 24/254) Есть формулы: cos (pi - x) = -cos x; tg (pi/2 + x) = -ctg x1 - cos x = 2sin^2 (x/2)Подставляем(1 + cos(pi - 4a))*tg(pi/2 + 2a) = (1 - cos(4a))*(-ctg(2a)) == -2sin^2(2a)*cos(2a)/sin(2a) = -2sin(2a)*cos(2a) = -sin(4a)5) tg a = 4/3; pi < a < 3pi/2; значит, sin a < 0; cos a < 0tg 2a = 2tg a/(1 - tg^2 a) = 2*(4/3) / (1 - 16/9) = (8/3) / (-7/9) = -24/7Из tg a = 4/3 следует: sin a = -4/5; cos a = -3/5cos (a + pi/6) = cos a*cos(pi/6) - sin a*sin(pi/6) == (-3/5)*(√3/2) - (-4/5)*(1/2) = -3√3/10 + 4/10 = (4 - 3√3)/10