Answer
$y=\frac{4\sqrt 2}{\pi}x-\sqrt 2$
Work Step by Step
The equation of the tangent line is of the form
$y=f(a)+f'(a)(x-a)$
Since $a=\frac{\pi}{4}$,
$f(a)=f(\frac{\pi}{4})=\frac{\sin\frac{\pi}{4}-\cos\frac{\pi}{4}}{\frac{\pi}{4}}=0$
$f'(\theta)=\frac{(\cos\theta+\sin\theta)\theta-(\sin\theta-\cos\theta)1}{\theta^{2}}$ (use quotient rule)
$f'(a)=f'(\frac{\pi}{4})=\frac{(\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2})\frac{\pi}{4}-0}{(\frac{\pi}{4})^{2}}=\frac{4\sqrt 2}{\pi}$
Therefore, the equation of the tangent line is
$y=0+\frac{4\sqrt 2}{\pi}(x-\frac{\pi}{4})$
Or $y=\frac{4\sqrt 2}{\pi}x-\sqrt 2$
