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

Question

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

Emma Aldridge · Accepted Answer

The Reqd. Slope#=2/(sqrt6+sqrt3-sqrt2)#.

Slope of tgt. to the curve # C : f(x)=cosx+sin(2x-pi/12)# at #x=pi/3# is #f'(pi/3)#. Hence, the slope of the normal at that pt. is #-1/(f'(pi/3))#. Now, #f(x)=cosx+sin(2x-pi/12)# #rArr f'(x)=-sinx+cos(2x-pi/12)*d/dx(2x-pi/12)# #=-sinx+2cos(2x-pi/12)# #rArr f'(pi/3)=-sin(pi/3)+2cos{2(pi/3)-pi/12}# #=-sqrt3/2+2cos(7pi/12)# Here, #2cos(7pi/12)=2cos(pi/3+pi/4)# #=2{cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)}# #=2{1/2*1/sqrt2-sqrt3/2*1/sqrt2}=2{(1-sqrt3)/(2sqrt2)}=(1-sqrt3)/sqrt2# Hence, #f'(pi/3)=-sqrt3/2+(1-sqrt3)/sqrt2=-sqrt3/2+(sqrt2-sqrt6)/2#, #=(sqrt2-sqrt3-sqrt6)/2# Therefore, the Reqd. Slope#=2/(sqrt6+sqrt3-sqrt2)#.

Paisley Johnson · Answer

undefined To find the slope of the line normal to the tangent line of $ f(x) = \cos(x) + \sin(2x - \frac{\pi}{12}) $ at $ x = \frac{\pi}{3} $, we first need to find the derivative of the function $ f(x) $ and evaluate it at $ x = \frac{\pi}{3} $ to find the slope of the tangent line. Then, we'll use the negative reciprocal of this slope to find the slope of the line normal to the tangent line.

First, find the derivative of $ f(x) $:
$$ f'(x) = -\sin(x) + 2\cos(2x - \frac{\pi}{12}) $$

Evaluate $ f'(x) $ at $ x = \frac{\pi}{3} $:
$$ f'(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) + 2\cos(2(\frac{\pi}{3}) - \frac{\pi}{12}) $$
$$ = -\sin(\frac{\pi}{3}) + 2\cos(\frac{2\pi}{3} - \frac{\pi}{12}) $$
$$ = -\frac{\sqrt{3}}{2} + 2\cos(\frac{3\pi}{4}) $$

Now, calculate $ \cos(\frac{3\pi}{4}) $:
$$ \cos(\frac{3\pi}{4}) = \frac{-\sqrt{2}}{2} $$

Substitute this value back into $ f'(\frac{\pi}{3}) $:
$$ = -\frac{\sqrt{3}}{2} + 2 \times \frac{-\sqrt{2}}{2} $$
$$ = -\frac{\sqrt{3}}{2} - \sqrt{2} $$

This is the slope of the tangent line. To find the slope of the line normal to the tangent line, take the negative reciprocal of this value:
$$ m_{normal} = -\frac{1}{-\frac{\sqrt{3}}{2} - \sqrt{2}} $$
$$ m_{normal} = \frac{1}{\frac{\sqrt{3}}{2} + \sqrt{2}} $$
$$ m_{normal} = \frac{2}{\sqrt{3} + 2\sqrt{2}} $$

This is the slope of the line normal to the tangent line of $ f(x) $ at $ x = \frac{\pi}{3} $.