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

Question

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

Elijah Abram · Accepted Answer

Slope #m_p=((sqrt(2+sqrt2)-2sqrt3)(sqrt2+10))/(-49)#

Slope #m_p=0.37651589912173# #f(x)=cos x+sin(2x-pi/12)" "#at #x=(5pi)/8# #f'(x)=-sin x+ 2*cos(2x-pi/12)# #f'((5pi)/8)=-sin((5pi)/8)+2*cos(2*((5pi)/8)-pi/12)# #f'((5pi)/8)=-cos (pi/8)+2*cos((7pi)/6)# #f'((5pi)/8)=-1/2sqrt(2+sqrt2)+2((-sqrt3)/2)# #f'((5pi)/8)=(-sqrt(2+sqrt2)-2sqrt3)/2# For the slope of the normal line #m_p=-1/m=-1/(f'((5pi)/8))=2/(sqrt(2+sqrt2)+2sqrt3)# #m_p=(2(sqrt(2+sqrt2)-2sqrt3))/(sqrt2-10)# #m_p=(2(sqrt(2+sqrt2)-2sqrt3)(sqrt2+10))/(-98)# #m_p=((sqrt(2+sqrt2)-2sqrt3)(sqrt2+10))/(-49)# God bless....I hope the explanation is useful.

Cameron Alexander · Answer

undefined To find the slope of the line normal to the tangent line of the function $ f(x) = \cos(x) + \sin(2x - \frac{\pi}{12}) $ at $ x = \frac{5\pi}{8} $, we first find the derivative of the function and evaluate it at $ x = \frac{5\pi}{8} $. Then, we find the negative reciprocal of this derivative to get the slope of the normal line.

The derivative of $ f(x) $ is $ f'(x) = -\sin(x) + 2\cos(2x - \frac{\pi}{12}) $.

Evaluate $ f'(x) $ at $ x = \frac{5\pi}{8} $:
$$ f'\left(\frac{5\pi}{8}\right) = -\sin\left(\frac{5\pi}{8}\right) + 2\cos\left(2\left(\frac{5\pi}{8}\right) - \frac{\pi}{12}\right) $$
$$ f'\left(\frac{5\pi}{8}\right) = -\sin\left(\frac{5\pi}{8}\right) + 2\cos\left(\frac{5\pi}{4} - \frac{\pi}{12}\right) $$
$$ f'\left(\frac{5\pi}{8}\right) = -\sin\left(\frac{5\pi}{8}\right) + 2\cos\left(\frac{15\pi}{12} - \frac{\pi}{12}\right) $$
$$ f'\left(\frac{5\pi}{8}\right) = -\sin\left(\frac{5\pi}{8}\right) + 2\cos\left(\frac{14\pi}{12}\right) $$
$$ f'\left(\frac{5\pi}{8}\right) = -\sin\left(\frac{5\pi}{8}\right) + 2\cos\left(\frac{7\pi}{6}\right) $$

Using the unit circle, $ \sin\left(\frac{5\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right) $ and $ \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} $.

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

The slope of the tangent line at $ x = \frac{5\pi}{8} $ is $ f'\left(\frac{5\pi}{8}\right) = -\sin\left(\frac{\pi}{8}\right) - \sqrt{3} $.

The slope of the normal line is the negative reciprocal of the slope of the tangent line:
$$ \text{Slope of normal line} = -\frac{1}{f'\left(\frac{5\pi}{8}\right)} = -\frac{1}{-\sin\left(\frac{\pi}{8}\right) - \sqrt{3}} $$