What is the equation of the normal line of #f(x)= tanx*sinx# at #x = pi/8#?

1 Answer

#y=-1.2033332887561x+0.63106054525314# OR
#y=-((14+9sqrt2)sqrt(2-sqrt(2)))/17*x+(pi(14+9sqrt(2))sqrt(2-sqrt2))/136#
#+((sqrt2-1)sqrt(2-sqrt2))/2#

Explanation:

Given # y=tan x * sin x #
and #x=pi/8#

Solve for the Ordinate #y_1# when Abscissa #x_1=pi/8#

#y_1=f(pi/8)=tan (pi/8) sin (pi/8)#

Take note from Trigonometry:

#Sin (A/2)=sqrt((1-cos A)/2)# and #cos (A/2)=sqrt((1+cos A)/2)#
#Tan (A/2)=(1-cos A)/Sin A#
Let #A=pi/4# So that
#Sin (pi/8)=sqrt(2-sqrt2)/2#
#Cos(pi/8)=sqrt(2+sqrt2)/2#
#Tan(pi/8)=sqrt(2)-1#

#y_1=f(pi/8)=tan (pi/8) sin (pi/8)#
#y_1=((sqrt(2)-1)*sqrt(2-sqrt2))/2#

The First derivative #f' (x)#:
#f' (x)=tan x*cos x+sin x*sec^2x#
Also
#f' (x)=sin x(1+sec^2x)#
And for the Normal Line
Slope #m=-1/(f' (pi/8))#

Slope #m=-1/(sin x(1+sec^2x)#

Slope #m=-1/((sqrt(2-sqrt2)/2)*(1+(2/(sqrt(2+sqrt2)))^2)#

Slope #m=-((14+9sqrt(2))sqrt(2-sqrt(2)))/17#

At this point, use now #(x_1, y_1)# and slope #m# to find the Normal Line.
Use Point-Slope Form #y-y_1=m(x-x_1)#

#y-((sqrt(2)-1)*sqrt(2-sqrt2))/2=#
#-((14+9sqrt(2))sqrt(2-sqrt(2)))/17 (x-pi/8)#

#y=-((14+9sqrt2)sqrt(2-sqrt(2)))/17*x+(pi(14+9sqrt(2))sqrt(2-sqrt2))/136#
#+((sqrt2-1)sqrt(2-sqrt2))/2#

OR using Decimal Values:

#y=-1.2033332887561x+0.63106054525314#

graph{y=-1.2033332887561x+0.63106054525314 [-2.5, 2.5, -1.25, 1.25]}

graph{y=tan x sinx [-2.5,2.5,-1.25,1.25]}