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

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Answer 1 · 2016-03-24T04:26:56

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#

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.