The function can be written as #f(x) = cscx + cscx = 2cscx#.
Start by finding the y-coordinate at the point of tangency.
#f(pi/6) = 2csc(pi/6) = 2/sin(pi/6) = 2/(1/2) = 4#
Differentiate.
#f(x) = 2/sinx#
#f'(x) = (0 * sinx - 2 * cosx)/(sinx)^2#
#f'(x) = (-2cosx)/sin^2x#
#f'(x) = -2cotxcscx#
You can find the slope of the tangent by inputting the point #x =a# into the derivative.
#f'(pi/6) = -2cot(pi/6)csc(pi/6)#
#f'(pi/6) = -2/(tan(pi/6)sin(pi/6))#
#f'(pi/6) = -2/(1/sqrt(3) * 1/2)#
#f'(pi/6) = -2/(1/(2sqrt(3))#
#f'(pi/6) = -4sqrt(3)#
The normal line is always perpendicular to the tangent. Therefore, its slope is given by the equation #m_"normal" = -1/m_"tangent"#.
#m_"normal" = -1/(-4sqrt(3))#
#m_"normal" = 1/(4sqrt(3))#
The equation of the normal is therefore:
#y - y_1 = m(x - x_1)#
#y - 4 = 1/(4sqrt(3))(x - pi/6)#
#y - 4= 1/(4sqrt(3))x - pi/(24sqrt(3))#
#y = 1/(4sqrt(3))x - (pi -96sqrt(3))/(24sqrt(3))#
Hopefully this helps!
