# What is the slope of the line normal to the tangent line of f(x) = xcotx+2xsin(x-pi/3)  at  x= (15pi)/8 ?

Aug 12, 2017

$- \frac{2}{15 \sqrt{3} \pi + 4}$

#### Explanation:

$\Rightarrow x \cot x + 2 x \sin \left(x - \frac{\pi}{3}\right)$

For reference :-
$\sin \left(\frac{15 \pi}{8}\right) = \frac{1}{2}$
$\cos \left(\frac{15 \pi}{8}\right) = - \frac{\sqrt{3}}{2}$
$\cos e c \left(\frac{15 \pi}{8}\right) = 2$
$\cot \left(\frac{15 \pi}{8}\right) = - \sqrt{3}$

As slope of any line is the derivative of the equation, I will differentiate the equation of tangent.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - x \cos e {c}^{2} x \cot x + 2 \left[\sin \left(x - \frac{\pi}{3}\right) + \cos \left(x - \frac{\pi}{3}\right)\right]$

clearly,
Here I can see the trigonometric formula of $\sin \left(A - B\right) \mathmr{and} \cos \left(A - B\right)$
Solving,
$\sin \left(x - \frac{\pi}{3}\right) + \cos \left(x - \frac{\pi}{3}\right)$
by above 2 formulas, I get,
$\left(\sin x + \cos x\right) + \sqrt{3} \left(\sin x - \cos x\right)$

So, at $x = \frac{15 \pi}{8}$,
The value of above equation becomes $2$

Now, at $x = \frac{15 \pi}{8}$,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{15 \sqrt{3} \pi + 4}{2} = {m}_{\text{1}}$

Now, when a line is normal to the tangent line, it means that the line is perpendicular to the tangent line.

Let the slope of perpendicular line be ${m}_{\text{2}}$

If the lines are perpendicular to each other then,
${m}_{\text{1"m_"2}} = - 1$

So, using above equation the slope of line normal to the tangent comes out to be,

${m}_{\text{2}} = - \frac{2}{15 \sqrt{3} \pi + 4}$

ENJOY MATHS !!!!!!!!