The slope of the line is
[tex]$y'\left(\frac{3\pi}4\right) = 9\cos\left(\frac{3\pi}4\right) = 9 \cdot \frac{-1}{\sqrt2} = \frac{-9}{\sqrt2} = \frac{-9\sqrt2}2,$[/tex]
And it must intersect the point [tex](3\pi/4,\,9\sqrt2/2)[/tex]. So the equation (point-slope form) is
[tex]$y - \frac{9\sqrt2}{2} = \frac{-9\sqrt2}2\left(x - \frac{3\pi}4\right)$[/tex]
How ugly.
