# What is the slope of the line normal to the tangent line of f(x) = cosx+sin(2x-pi/12)  at  x= pi/3 ?

Aug 20, 2016

The Reqd. Slope$= \frac{2}{\sqrt{6} + \sqrt{3} - \sqrt{2}}$.

#### Explanation:

Slope of tgt. to the curve $C : f \left(x\right) = \cos x + \sin \left(2 x - \frac{\pi}{12}\right)$ at $x = \frac{\pi}{3}$ is $f ' \left(\frac{\pi}{3}\right)$.

Hence, the slope of the normal at that pt. is $- \frac{1}{f ' \left(\frac{\pi}{3}\right)}$.

Now, $f \left(x\right) = \cos x + \sin \left(2 x - \frac{\pi}{12}\right)$

$\Rightarrow f ' \left(x\right) = - \sin x + \cos \left(2 x - \frac{\pi}{12}\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x - \frac{\pi}{12}\right)$

$= - \sin x + 2 \cos \left(2 x - \frac{\pi}{12}\right)$

$\Rightarrow f ' \left(\frac{\pi}{3}\right) = - \sin \left(\frac{\pi}{3}\right) + 2 \cos \left\{2 \left(\frac{\pi}{3}\right) - \frac{\pi}{12}\right\}$

$= - \frac{\sqrt{3}}{2} + 2 \cos \left(7 \frac{\pi}{12}\right)$

Here, $2 \cos \left(7 \frac{\pi}{12}\right) = 2 \cos \left(\frac{\pi}{3} + \frac{\pi}{4}\right)$

$= 2 \left\{\cos \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) - \sin \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)\right\}$

$= 2 \left\{\frac{1}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}\right\} = 2 \left\{\frac{1 - \sqrt{3}}{2 \sqrt{2}}\right\} = \frac{1 - \sqrt{3}}{\sqrt{2}}$

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

$= \frac{\sqrt{2} - \sqrt{3} - \sqrt{6}}{2}$

Therefore, the Reqd. Slope$= \frac{2}{\sqrt{6} + \sqrt{3} - \sqrt{2}}$.