The equation of the tangent line may be identified using the first derivative of the function which gives us its slope.
[tex]\begin{gathered} y=\cos3x \\ \\ y^{\prime}=-3\sin3x \\ \\ m=-3\sin3x \\ \\ m=-3\sin3(\frac{\pi}{4}) \\ \\ m=-3(\frac{\sqrt{2}}{2}) \\ \\ m=-\frac{3\sqrt{2}}{2} \end{gathered}[/tex]
The tangent line passes through (/4, -√2/2) so we can solve for the y-intercept, b.
[tex]\begin{gathered} y=mx+b \\ \\ -\frac{\sqrt{2}}{2}=-\frac{3\sqrt{2}}{2}(\frac{\pi}{4})+b \\ \\ -\frac{\sqrt{2}}{2}=-\frac{3\pi\sqrt{2}}{8}+b \\ \\ b=-\frac{\sqrt{2}}{2}+\frac{3\pi\sqrt{2}}{8} \\ \\ b=\frac{-4\sqrt{2}}{8}+\frac{3\pi\sqrt{2}}{8} \\ \\ b=\frac{(3\pi-4)\sqrt{2}}{8} \end{gathered}[/tex]
So the equation of the tangent line is:
[tex]y=-\frac{3\sqrt{2}}{2}x+\frac{(3\pi-4)\sqrt{2}}{8}[/tex]
