Question

Дано: cosa= 1/√2, cosb = 1/3, a и b - углы 4 четверти. Вычислите: а) sin (a-b); б) cos (a+b); в) tg (a+b)​

Answer 1

Ответ:

a) sin(a-b) = sin(a)cos(b) - cos(a)sin(b) = cosa*cosb - cosb*cosa = (1/√2)*(1/3) - (1/3)*(1/√2) = (1 - √2/6)/2 = (1 - √2/3)/(2√2) = 0.154

b) cos(a+b) = cos(a)cos(b) + sin(a)sin(b) = cosa*cosb + cosb*cosa = (1/√2)*(1/3) + (1/3)*(1/√2) = (1 + √2/6)/2 = (1 + √2/3)/(2√2) = 0.846

в) tg(a+b) = tg(a) + tg(b) = (1/√2)/(1/3) + (1/3)/(1/√2) = (3*√2 + 1)/(3 - √2) = (3*√2 + 1)/(√2 - 1) ~= 2.29