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`?

Answer 1

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

Explanation:

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)`.