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]}
