What is the slope of the line normal to the tangent line of f(x) = secx+sin(2x-(3pi)/8) at x= (11pi)/8 ?

1 Answer

The slope of the line normal to the tangent line
m=1/((1+sqrt(2)/2)sqrt(2+sqrt2)+((3sqrt2)/2+1)sqrt(2-sqrt2)
m=0.18039870004873

Explanation:

From the given:
y=sec x+ sin (2x-(3pi)/8) at " "x=(11pi)/8

Take the first derivative y'

y'=sec x*tan x *(dx)/(dx)+ cos (2x-(3pi)/8)(2)(dx)/(dx)

Using " "x=(11pi)/8

Take note: that by color(Blue)("Half-Angle formulas"), the following are obtained
sec ((11pi)/8)=-sqrt(2+sqrt2)-sqrt(2-sqrt2)

tan ((11pi)/8)=sqrt2+1

and

2*cos (2x-(3pi)/8)=2*cos ((19pi)/8)
=2*(sqrt2/4)(sqrt(2+sqrt2)-sqrt(2-sqrt2))

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
continuation

y'=(-sqrt(2+sqrt2)-sqrt(2-sqrt2))(sqrt2+1)
+2*(sqrt2/4)(sqrt(2+sqrt2)-sqrt(2-sqrt2))

y'=-(sqrt2+1)sqrt(2+sqrt2)-(sqrt2+1)sqrt(2-sqrt2)
+(sqrt2)/2*sqrt(2+sqrt2)-sqrt2/2*sqrt(2-sqrt2)

further simplification

y'=(-1-sqrt2/2)sqrt(2+sqrt2)+((-3sqrt2)/2-1)sqrt(2-sqrt2)

For the normal line: m=(-1)/(y')

m=(-1)/((-1-sqrt2/2)sqrt(2+sqrt2)+((-3sqrt2)/2-1)sqrt(2-sqrt2))

m=1/((1+sqrt2/2)sqrt(2+sqrt2)+((3sqrt2)/2+1)sqrt(2-sqrt2))

m=0.180398700048733

God bless....I hope the explanation is useful.