What is the slope of the line normal to the tangent line of $f (x) = {cot}^{2} x + sin (x - \frac{π}{4})$ $f(x) = cot^2x+sin(x-pi/4)$ at $x = \frac{5 π}{6}$ $x= (5pi)/6$ ?

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What is the slope of the line normal to the tangent line of $f (x) = {cot}^{2} x + sin (x - \frac{π}{4})$ $f(x) = cot^2x+sin(x-pi/4)$ at $x = \frac{5 π}{6}$ $x= (5pi)/6$ ?

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Answer 1 · 2018-06-02T17:52:17

the slope of the normal line is given by
$\frac{1}{586753} (\sqrt{6} + \sqrt{2} - 32 \cdot \sqrt{3}) \cdot (749 + 48 \cdot \sqrt{2})$ $1/586753(sqrt(6)+sqrt(2)-32*sqrt(3))*(749+48*sqrt(2))$

Explanation:

$f'(x)=cos(pi/4-x)-2cot(x)*csc^2(x)$
so
$f'(5*pi/6)=1/4*(sqrt(2)+32sqrt(3)-sqrt(6))$
so the slope of the normal line is given by
$m_N=-1/(1/4*(sqrt(2)+32*sqrt(3)-sqrt(6))$

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What is the slope of the line normal to the tangent line of $f (x) = {cot}^{2} x + sin (x - \frac{π}{4})$ $f(x) = cot^2x+sin(x-pi/4)$ at $x = \frac{5 π}{6}$ $x= (5pi)/6$ ?

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What is the slope of the line normal to the tangent line of f(x) = cot^2x+sin(x-pi/4) f(x)=cot2x+sin(x−π4)f(x) = cot^2x+sin(x-pi/4) at x= (5pi)/6 x=5π6 x= (5pi)/6 ?

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What is the slope of the line normal to the tangent line of $f (x) = {cot}^{2} x + sin (x - \frac{π}{4})$ $f(x) = cot^2x+sin(x-pi/4)$ at $x = \frac{5 π}{6}$ $x= (5pi)/6$ ?