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

Jan 6, 2016

${m}_{n} = \frac{\sqrt{2}}{2}$

Explanation:

Using the product rule, we can find the derivative or the "slope function" of $f \left(x\right)$.
$f ' \left(x\right) = - 2 \sin \left(2 x\right) \cdot \sin \left(2 x - \frac{\pi}{12}\right) + \cos \left(2 x\right) \cdot 2 \cos \left(2 x - \frac{\pi}{12}\right)$
$f ' \left(x\right) = 2 \left(\cos \left(2 x\right) \cdot \cos \left(2 x - \frac{\pi}{12}\right) - \sin \left(x\right) \cdot \sin \left(2 x - \frac{\pi}{12}\right)\right)$.

Just to make it a bit neater, we can use the expansion formula for $\cos \left(a + b\right) = \cos \left(a\right) \cos \left(b\right) - \sin \left(a\right) \sin \left(b\right)$ to rewrite $f ' \left(x\right)$.

$f ' \left(x\right) = 2 \cos \left(4 x - \frac{\pi}{12}\right)$.

Hence, $f ' \left(\frac{\pi}{3}\right) = 2 \cos \left(4 \frac{\pi}{3} - \frac{\pi}{12}\right)$

$f ' \left(\frac{\pi}{3}\right) = 2 \cos \left(15 \frac{\pi}{12}\right)$
$f ' \left(\frac{\pi}{3}\right) = 2 \cos \left(5 \frac{\pi}{4}\right)$
$f ' \left(\frac{\pi}{3}\right) = - \sqrt{2}$

Now, that's the slope of the tangent, the slope of the normal would be the negative reciprocal of that, in other words, ${m}_{n} \cdot \left(- \sqrt{2}\right) = - 1$.

Hence, ${m}_{n} = \frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$.