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`?

Answer 1

`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`.