# What is the equation of the normal line of f(x)=1/sinx+cscx at x=pi/6?

Jan 4, 2017

$y = \frac{1}{4 \sqrt{3}} x - \frac{\pi - 96 \sqrt{3}}{24 \sqrt{3}}$

#### Explanation:

The function can be written as $f \left(x\right) = \csc x + \csc x = 2 \csc x$.

Start by finding the y-coordinate at the point of tangency.

$f \left(\frac{\pi}{6}\right) = 2 \csc \left(\frac{\pi}{6}\right) = \frac{2}{\sin} \left(\frac{\pi}{6}\right) = \frac{2}{\frac{1}{2}} = 4$

Differentiate.

$f \left(x\right) = \frac{2}{\sin} x$

$f ' \left(x\right) = \frac{0 \cdot \sin x - 2 \cdot \cos x}{\sin x} ^ 2$

$f ' \left(x\right) = \frac{- 2 \cos x}{\sin} ^ 2 x$

$f ' \left(x\right) = - 2 \cot x \csc x$

You can find the slope of the tangent by inputting the point $x = a$ into the derivative.

$f ' \left(\frac{\pi}{6}\right) = - 2 \cot \left(\frac{\pi}{6}\right) \csc \left(\frac{\pi}{6}\right)$

$f ' \left(\frac{\pi}{6}\right) = - \frac{2}{\tan \left(\frac{\pi}{6}\right) \sin \left(\frac{\pi}{6}\right)}$

$f ' \left(\frac{\pi}{6}\right) = - \frac{2}{\frac{1}{\sqrt{3}} \cdot \frac{1}{2}}$

f'(pi/6) = -2/(1/(2sqrt(3))

$f ' \left(\frac{\pi}{6}\right) = - 4 \sqrt{3}$

The normal line is always perpendicular to the tangent. Therefore, its slope is given by the equation ${m}_{\text{normal" = -1/m_"tangent}}$.

${m}_{\text{normal}} = - \frac{1}{- 4 \sqrt{3}}$

${m}_{\text{normal}} = \frac{1}{4 \sqrt{3}}$

The equation of the normal is therefore:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 4 = \frac{1}{4 \sqrt{3}} \left(x - \frac{\pi}{6}\right)$

$y - 4 = \frac{1}{4 \sqrt{3}} x - \frac{\pi}{24 \sqrt{3}}$

$y = \frac{1}{4 \sqrt{3}} x - \frac{\pi - 96 \sqrt{3}}{24 \sqrt{3}}$

Hopefully this helps!