What is the equation of the normal line of $f (x) = sin x$ $f(x)= sinx$ at $x = \frac{π}{8}$ $x = pi/8$ ?

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Answer 1 · 2017-02-16T16:51:40

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

f at $x = \frac{π}{8}$ $x = pi/8$ is $sin (\frac{π}{8}) = \sqrt{\frac{1}{2} (1 - cos (\frac{π}{4}))}$ $sin(pi/8)=sqrt(1/2(1-cos(pi/4))$

$= \frac{1}{2} \sqrt{1 - \frac{1}{\sqrt{2}}} = 0.5837$ $=1/2sqrt(1-1/sqrt2)=0.5837$ , nearly.

So, the foot of the normal P is $(sin (\frac{π}{8}), \frac{π}{8}) = (0.3927, 0.5837)$ $(sin(pi/8), pi/8)=( 0.3927, 0.5837)$

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

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]}

What is the equation of the normal line of f(x)= sinxf(x)=sinxf(x)= sinx at x = pi/8x=π8x = pi/8?