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 #?

Question

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  #?

Ezekiel Alexander · Accepted Answer

#m_n = sqrt(2)/2# Using the product rule, we can find the derivative or the "slope function" of #f(x)#. #f'(x) = -2sin(2x) * sin(2x - pi/12) + cos(2x) * 2cos(2x - pi/12)# #f'(x) = 2(cos(2x) * cos(2x - pi/12) - sin(x) * sin(2x - pi/12))#. Just to make it a bit neater, we can use the expansion formula for #cos(a+b) = cos(a)cos(b) - sin(a)sin(b)# to rewrite #f'(x)#. #f'(x) = 2cos(4x - pi/12)#. Hence, #f'(pi/3) = 2cos(4pi/3 - pi/12)# #f'(pi/3) = 2cos(15pi/12)# #f'(pi/3) = 2cos(5pi/4)# #f'(pi/3) = -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 * (-sqrt(2)) = -1#. Hence, #m_n = 1/sqrt(2)# or #sqrt(2)/2#.

William Abdalla · Answer

undefined To find the slope of the line normal to the tangent line of $ f(x) = \cos(2x) \cdot \sin\left(2x - \frac{\pi}{12}\right) $ 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 get the slope of the tangent line. Then, we find the negative reciprocal of this slope to get the slope of the normal line. Finally, we can plug in the value of $ x = \frac{\pi}{3} $ into the function to find the corresponding y-coordinate on the curve.