Answer
$y=-\dfrac{3\sqrt{2}}{2}x+\dfrac{3\sqrt{2}\pi -4\sqrt{2}}{8}.$
Work Step by Step
$u=3x$; $\dfrac{du}{dx}=3$
$\dfrac{dy}{du}=-\sin{u}$
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=-3\sin{3x}$
$ y' $ evaluated at $(\dfrac{\pi}{4}, -\dfrac{\sqrt{2}}{2})\rightarrow y'= -3\sin{\dfrac{3\pi}{4}}=-\dfrac{3\sqrt{2}}{2}.$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y+\dfrac{\sqrt{2}}{2})=-\dfrac{3\sqrt{2}}{2}(x-\dfrac{\pi}{4})\rightarrow y=-\dfrac{3\sqrt{2}}{2}x+\dfrac{3\sqrt{2}\pi -4\sqrt{2}}{8}.$
A graphing calculator and a computer algebra system have been used to confirm these results.
