Answer
$y-1 = -3(x-\frac{3\pi}{4})$
Work Step by Step
We first solve for $f'(x)$:
$f'(x) = Mcos(x)-Nsin(x)$
Thus, we have:
$f(\frac{\pi}{4}) = Msin(\frac{\pi}{4}) + Ncos(\frac{\pi}{4}) = 3$
$f'(\frac{\pi}{4}) = Mcos(\frac{\pi}{4}) - Nsin(\frac{\pi}{4}) = 1$
$M\frac{\sqrt{2}}{2} + N\frac{\sqrt{2}}{2} = 3$
$M\frac{\sqrt{2}}{2} - N\frac{\sqrt{2}}{2}= 1$
Thus, adding the two equations, we get:
$\sqrt{2}M = 4$
$M = 2\sqrt{2}$
Subtracting the two equations, we get:
$\sqrt{2}N = 2$
$N = \sqrt{2}$
Thus, $f(x) = 2\sqrt{2}sin(x) + \sqrt{2}cos(x)$ and $f'(x) = 2\sqrt{2}cos(x) - \sqrt{2}sin(x)$
Therefore,
$f(\frac{3\pi}{4}) = 2\sqrt{2}sin(\frac{3\pi}{4}) +\sqrt{2}cos(\frac{3\pi}{4}) = 1$
$f'(\frac{3\pi}{4}) = 2\sqrt{2}cos(\frac{3\pi}{4}) - \sqrt{2}sin(\frac{3\pi}{4}) = -3$
The equation for the tangent line is then:
$y-1 = -3(x-\frac{3\pi}{4})$