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!
