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 #?

1 Answer
Jan 6, 2016

#m_n = sqrt(2)/2#

Explanation:

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#.