Вопрос пользователя:
64*sin(x)*cos(4x)*cos(8x)*Cos(16x)*cos(32x)*cos(x)=1
Илюха отвечает:
найдём производные
f'(x) = (sin(4x))’ – (cos(2x))’ = 4cos(4x) + 2sin(2x)
g'(x) = (cos^2(2x))’= 2cos(2x) * (-sin(2x))*2 = -2sin(4x)
y'(x) = -sin(x)/(1-cos(x)) – (1+cos(x))*sin(x)/(1-cos(x))^2 =
= -sin(x)/(1-cos(x))^2 * (1 – cos(x) + 1 + cos(x)) = -2sin(x)/(1-cos(x))^2
y”(x) = -2cos(x)/(1-cos(x)^2 -2sin(x) * sin(x) * (-2)/(1-cos(x))^3 =
= (-2cos(x)*(1-cos(x) + 4sin^2(x))/(1-cos(x))^3 = 2(2+cos(x))/(1-cos(x))^2
корень – sqrt
y”(pi/4) = 2*(2 + sqrt(2)/2)/(1 – sqrt(2)/2)^2 = (4 + sqrt(2))/(1 + 1/2 – sqrt(2)) =
= (4 + sqrt(2)) / (3/2 – sqrt(2)) = (4 + sqrt(2))*(1.5 + sqrt(2)) / (2.25 – 2) =
= (6 + 1.5sqrt(2) + 4sqrt(2) + 2)/ 0.25 = 32 + 22sqrt(2)