What is the equation of the normal line of f(x)=3ln⁡sin⁡2x at x=π8?

Question

szalbierzfytg · Accepted Answer

While slope of tangent at curve f(x) at x=a is given by f'(a), as lope of normal
at x=a is

\frac{- 1}{f^{'} (a)}

as normal is perpendicular to the tangent.
Both pass through the point (a, f(a)) on the curve.
Here we have

f (x) = 3 \ln \sin 2 x

and hence

f^{'} (x) = \frac{3}{\sin 2 x} \times 2 \cos 2 x = 6 \cot 2 x

and hence we are seeking a normal at

(\frac{π}{8}, 6 \cot (\frac{π}{4}))

or

(\frac{π}{8}, 6)

and as slope of normal is

\frac{- 1}{6 \cot (\frac{π}{4})} = - \frac{1}{6}

and as equation of line passing through

(x_{1}, y_{1})

and having a slope m is

y - y_{1} = m (x - x_{1})

and hence equation of normal is

y - 6 = \frac{- 1}{6} (x - \frac{π}{8})

or

6 y - 36 = - x + \frac{π}{8}

or

x + 6 y - 36 - \frac{π}{8} = 0

What is the equation of the normal line