What is the slope of the line normal to the tangent line of $f (x) = x cot x - {sin}^{2} (x - π)$ $f(x) = xcotx-sin^2(x-pi)$ at $x = \frac{π}{12}$ $x= (pi)/12$ ?

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What is the slope of the line normal to the tangent line of $f (x) = x cot x - {sin}^{2} (x - π)$ $f(x) = xcotx-sin^2(x-pi)$ at $x = \frac{π}{12}$ $x= (pi)/12$ ?

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Answer 1 · 2018-07-23T15:48:18

$y = (\frac{3}{2} + \sqrt{3} - \frac{2}{3} \cdot π - \frac{π}{\sqrt{3}}) x + \frac{1}{72} (- 36 + 18 \cdot \sqrt{3} + 3 π + 4 π^{2} + 2 \sqrt{3} \cdot π^{2})$ $y=(3/2+sqrt(3)-2/3*pi-pi/sqrt(3))x+1/72(-36+18*sqrt(3)+3pi+4pi^2+2sqrt(3)*pi^2)$

Explanation:

given is

$f (x) = x cot (x) - {sin}^{2} (x - π)$ $f(x)=xcot(x)-sin^2(x-pi)$ then we get by the sum , the chain and the product rule:

$f'(x)=cot(x)-xcsc^2(x)-2sin(x)cos(x)$

now we compute

$f'(pi/12)=3/2+sqrt(3)-2/3*pi-pi/sqrt(3)$

and

$f(pi/12)=1/12(2+sqrt(3))pi-1/8*(sqrt(3)-1)^2$

The Tangent line is given by the equation

$y=mx+n$

$m=f'(pi/12)$

plugging $x=pi/12$ and $f(pi/12)$ in the given equation we get

$n=1/72(-36+18sqrt(3)+3pi+4pi^2+2sqrt(3)pi^2)$

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What is the slope of the line normal to the tangent line of $f (x) = x cot x - {sin}^{2} (x - π)$ $f(x) = xcotx-sin^2(x-pi)$ at $x = \frac{π}{12}$ $x= (pi)/12$ ?

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What is the slope of the line normal to the tangent line of f(x) = xcotx-sin^2(x-pi) f(x)=xcotx−sin2(x−π)f(x) = xcotx-sin^2(x-pi) at x= (pi)/12 x=π12 x= (pi)/12 ?

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Explanation:

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What is the slope of the line normal to the tangent line of $f (x) = x cot x - {sin}^{2} (x - π)$ $f(x) = xcotx-sin^2(x-pi)$ at $x = \frac{π}{12}$ $x= (pi)/12$ ?