What is the slope of the line normal to the tangent line of $f (x) = cos x \cdot sin (2 x - \frac{π}{6})$ $f(x) = cosx*sin(2x-pi/6)$ at $x = \frac{π}{3}$ $x= pi/3$ ?

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Answer 1 · 2017-07-25T19:16:17

$m = \frac{2}{\sqrt{3}} = \frac{2}{3} \sqrt{3}$ $m=2/sqrt3=2/3sqrt3$

Let's first find the derivative of the function using the product rule :

$f'(x)=-sinxsin(2x-pi/6)+2cosxcos(2x-pi/6)$

The slope of the tangent line is $f'(pi/3)$

$f'(pi/3)=-sin(pi/3)sin((2pi)/3-pi/6)+2cos(pi/3)cos((2pi)/3-pi/6)=$

$-sqrt3/2sin(pi/2)+2*1/2cos(pi/2)=-sqrt3/2$

So now the slope of the normal line to the tangent is :

$m=-1/(f'(pi/3))=2/sqrt3=2/3sqrt3$

What is the slope of the line normal to the tangent line of f(x) = cosx*sin(2x-pi/6) f(x)=cosx⋅sin(2x−π6)f(x) = cosx*sin(2x-pi/6) at x= pi/3 x=π3 x= pi/3 ?