# What is the slope of the line normal to the tangent line of f(x) = cosx+sin(2x-pi/12)  at  x= (5pi)/8 ?

Slope ${m}_{p} = \frac{\left(\sqrt{2 + \sqrt{2}} - 2 \sqrt{3}\right) \left(\sqrt{2} + 10\right)}{- 49}$

Slope ${m}_{p} = 0.37651589912173$

#### Explanation:

$f \left(x\right) = \cos x + \sin \left(2 x - \frac{\pi}{12}\right) \text{ }$at $x = \frac{5 \pi}{8}$

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

$f ' \left(\frac{5 \pi}{8}\right) = - \sin \left(\frac{5 \pi}{8}\right) + 2 \cdot \cos \left(2 \cdot \left(\frac{5 \pi}{8}\right) - \frac{\pi}{12}\right)$

$f ' \left(\frac{5 \pi}{8}\right) = - \cos \left(\frac{\pi}{8}\right) + 2 \cdot \cos \left(\frac{7 \pi}{6}\right)$

$f ' \left(\frac{5 \pi}{8}\right) = - \frac{1}{2} \sqrt{2 + \sqrt{2}} + 2 \left(\frac{- \sqrt{3}}{2}\right)$

$f ' \left(\frac{5 \pi}{8}\right) = \frac{- \sqrt{2 + \sqrt{2}} - 2 \sqrt{3}}{2}$

For the slope of the normal line

${m}_{p} = - \frac{1}{m} = - \frac{1}{f ' \left(\frac{5 \pi}{8}\right)} = \frac{2}{\sqrt{2 + \sqrt{2}} + 2 \sqrt{3}}$

${m}_{p} = \frac{2 \left(\sqrt{2 + \sqrt{2}} - 2 \sqrt{3}\right)}{\sqrt{2} - 10}$

${m}_{p} = \frac{2 \left(\sqrt{2 + \sqrt{2}} - 2 \sqrt{3}\right) \left(\sqrt{2} + 10\right)}{- 98}$

${m}_{p} = \frac{\left(\sqrt{2 + \sqrt{2}} - 2 \sqrt{3}\right) \left(\sqrt{2} + 10\right)}{- 49}$

God bless....I hope the explanation is useful.