Find the first derivative #-> #the slope function:
#f'(x) = cos x - (-sin 2x)(2) = cos x + 2 sin 2x#
Find the slope of the tangent line: #m_t = f'(pi/8) = cos(pi/8) + 2 sin(pi/4) = sqrt(2+sqrt(2))/2 + (2sqrt(2))/2#
#m_t = (sqrt(2+sqrt(2)) + 2sqrt(2))/2#
The normal line is the perpendicular line which the negative reciprocal of the tangent line for its slope:
#m_n = -2/(sqrt(2+sqrt(2)) + 2sqrt(2))#
#y_n -y_1= m_n(x - x_1)#
#f(pi/8) = sin(pi/8) - cos(pi/4) = (sqrt(2-sqrt(2)))/2 - sqrt(2)/2 = ((sqrt(2-sqrt(2)))- sqrt(2))/2#
#y_n -y_1= m_n(x - x_1):#
# y_n-((sqrt(2-sqrt(2)))- sqrt(2))/2 = -2/(sqrt(2+sqrt(2)) + 2sqrt(2))(x-pi/8)#
#y_n = -(2x)/(sqrt(2+sqrt(2)) + 2sqrt(2)) + pi/(4(sqrt(2+sqrt(2)) + 2sqrt(2))) + ((sqrt(2-sqrt(2)))- sqrt(2))/2#
