What is the slope of the line normal to the tangent line of #f(x) = tanx+sin(x-pi/4) # at # x= (5pi)/8 #?

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Answer 1 · 2016-12-29T03:23:48

#-0.1290#, nearly.

At #x=5/8pi,

y= f(5/8pi)=tan(5/8pi)+sin(5/8pi-pi/4)#

#=tan(pi/2+pi/8)+sin(-pi/8)#

#=-cot(pi/8)-sin(pi/8)=-2.797#, nearly

The slope of the normal at this point

#P(5/8pi, 2.0315)=P(1.9635,2.0315)#

is -1/y/ at P

#=-1/(tan x+sin(x-pi/4))' # at #x =5/8pi#

#=-1/(sec^2(5/8pi)+cos(pi/8)#

#=-1/(csc^2(pi/8)+cos(pi/8))#

#=-0.1290#, nearly.

The equation to the normal at P(1.9635,2.0315)# is

y-2.0315=-0.1290(x-1.9635)#. Simplifying,

#y=-0.1290 x+2.057#