sin⁴(π/16) + sin⁴(3π/16) + sin⁴(5π/16) + sin⁴(7π/16) = (1 - cos(π/8))²/4 +
+ (1 - cos(3π/8))²/4 + (1 - cos(5π/8))²/4 + (1 - cos(7π/8))²/4 = (1/4) •
• ( 1 - 2cos(π/8) + cos²(π/8) + 1 - 2cos(3π/8) + cos²(3π/8) + 1 - 2cos(5π/8) + cos²(5π/8) + 1 - 2cos(7π/8) + cos²(7π/8) ) = (1/4) • ( 4 - 2•( cos(π/8) + cos(3π/8) + cos(5π/8) + cos(7π/8) ) + ( cos²(π/8) + cos²(3π/8) + cos²(5π/8) + cos²(7π/8) ) ) = (1/4) • ( 4 - 2•( 2•cos(π/2)•cos(-3π/8) + 2•cos(π/2)•cos(-π/8) ) + ( cos²(π/8) + cos²(3π/8) + cos²(5π/8) + cos²(7π/8) ) ) = 1 + (1/4)•( cos²(π/8) + cos²(3π/8) + cos²(5π/8) + cos²(7π/8) ) = 1 + (1/4)•( ( cos(π/8) + cos(7π/8) )² + ( cos(3π/8) + cos(5π/8) )² - 2•cos(π/8)•cos(7π/8) - 2•cos(3π/8)•cos(5π/8) ) =
= 1 - (1/4)•( cosπ + cos(-3π/4) + cosπ + cos(-π/4) ) = 1 - (1/4)•( - 2 - (√2/2) + (√2/2) ) = 1 - (1/4)•(-2) = 1 + 0,5 = 1,5
ответ: 1,5
q2=(m² +2n²) второе число
q1*q2=(a² +2b²) *(m²+2n²) =a²m² +2m²b²+2a²n²+4b²n²=
=(am)²+(2bn)² +2((mb)²+(an)²)
До полного квадрата не хватает выражения
в первой скобке 4ambn добавляешь и вычитаешь его
(am)²+(2bn)²+4ambn-4ambn +2((mb)²+(an)²)=(am+2bn)²-4ambn + 2((mb)²+(an)²)
внесем -4ambn в скобку 2((mb)²+(an)²)
(am+2bn)²+2(mb)²+(an)²-2ambn)=(am+2bn)²+2(mb-an)²
произведем замену x=(am+2bn) y=(mb-an)
получим q1*q2=x²+2y² что и требовалось доказать