# What is the slope of the line normal to the tangent line of f(x) = cosx-cos^2x +1  at  x= (17pi)/12 ?

Slope of the normal to the tangent line ${m}_{n}$

$= \frac{- 1}{\frac{1}{4} \left(\sqrt{6 + 3 \sqrt{3}} + \sqrt{2 - \sqrt{3}}\right) + \frac{1}{8} \left(\sqrt{21 + 12 \sqrt{3}} - \sqrt{21 - 12 \sqrt{3}} - 2\right)}$

${m}_{n} = - 0.68216275480416$

#### Explanation:

From the given equation $f \left(x\right) = \cos x - {\cos}^{2} x + 1$
Obtain the first derivative

$f \left(x\right) = \cos x - {\cos}^{2} x + 1$

$f ' \left(x\right) = - \sin x - 2 \cos x \cdot \left(- \sin x\right) + 0$

$f ' \left(x\right) = - \sin x + 2 \sin x \cos x$

Slope $m = f ' \left(\frac{17 \pi}{12}\right)$

$m = f ' \left(\frac{17 \pi}{12}\right) = - \sin \left(\frac{17 \pi}{12}\right) + 2 \sin \left(\frac{17 \pi}{12}\right) \cos \left(\frac{17 \pi}{12}\right)$

Using trigonometric formulas like half-angle formulas and sum/difference formulas

$m = \frac{1}{4} \left(\sqrt{6 + 3 \sqrt{3}} + \sqrt{2 - \sqrt{3}}\right) + \frac{1}{8} \left(\sqrt{21 + 12 \sqrt{3}} - \sqrt{21 - 12 \sqrt{3}} - 2\right)$

Now, we have to solve for the slope of the normal line ${m}_{n}$

${m}_{n} = \frac{- 1}{m}$

${m}_{n} =$

$= \frac{- 1}{\frac{1}{4} \left(\sqrt{6 + 3 \sqrt{3}} + \sqrt{2 - \sqrt{3}}\right) + \frac{1}{8} \left(\sqrt{21 + 12 \sqrt{3}} - \sqrt{21 - 12 \sqrt{3}} - 2\right)}$

${m}_{n} = - 0.68216275480416$

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