What is the equation of the normal line of $f (x) = \frac{tan (2 x)}{cot x}$ $f(x)= tan(2x)/cotx$ at $x = \frac{π}{8}$ $x = pi/8$ ?

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Answer 1 · 2016-09-23T11:54:09

$- 2 (\sqrt{2} + 1) (2 - \sqrt{2}) (y - \frac{2 - \sqrt{2}}{2}) = x - \frac{π}{8}$ $-2(sqrt2+1)(2-sqrt2)(y-(2-sqrt2)/2)=x-pi/8$

equation of the normal: $-y'_0(y-y_0)=x-x_0$

$y'=((2/cos^2(2x))*cotx+(tan2x)/sin^2x)/cot^2x$

$y'_0=((2/cos^2(pi/4))*cot(pi/8)+(tan(pi/4))/sin^2(pi/8))/cot^2(pi/8)=$

$=(4cos(pi/8)sin(pi/8)+1)/cos^2(pi/8)=$

$=(2sin(pi/4)+1)/((cos(pi/4)+1)/2)=$

$=4*(sqrt2+1)/(sqrt2+2)$

$y_0=tan(pi/8)=1-sqrt2/2$

$-2(sqrt2+1)(2-sqrt2)(y-(2-sqrt2)/2)=x-pi/8$

What is the equation of the normal line of f(x)= tan(2x)/cotxf(x)=tan(2x)cotxf(x)= tan(2x)/cotx at x = pi/8x=π8x = pi/8?