Question

sin^2 p/12 * cos^2 p/12 (через понижение степени, но как о.о)

Answer 1

sin²x = (1-cos2x)/2

cos²x = (1+cos2x)/2

sin²π/12 * cos²π/12 = (1-cosπ/6)*(1+cosπ/6)/2*2 = (1 - cos²π/6)/4 = (1 - (√3/2)²)/4 = 

= (1 - 3/4)/4 = (1/4)/4 = 1/16

Ответ: 1/16