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

Answer 1

Ezra Adams

1.082x+y-1.009, nearly.. See the tangent-inclusive Socratic graph.

f at #x = pi/8 # is #sin(pi/8)=sqrt(1/2(1-cos(pi/4))#

#=1/2sqrt(1-1/sqrt2)=0.5837#, nearly.

So, the foot of the normal P is #(sin(pi/8), pi/8)=( 0.3927, 0.5837)#

#f'=cosx=cos(pi/8)=sqrt(1/2(1+cos(pi/4))) = sqrt(1/2(1+1/sqrt2))=0.9239#,

nearly. at P.

The slope of the normal at P is #-1/(f')=-1/0.9239=-1.082#, nearly.

Now, the equation to the normal, at P( (0.5837, 0.3927 ), is

#y-0,5837=-1.082(x-0.3927)#, giving

1.082x+y-1.009, nearly.

graph{(sinx-y)(1.082x+y-1.01)=0 [-10, 10, -5, 5]}

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Answer 2

Ezekiel Alexander

The equation of the normal line of f(x) = sinx at x = pi/8 is y = -8x + 1.

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Answer from HIX Tutor

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