Answer
$y=\frac{1}{2}x+\frac{3}{2}$
Work Step by Step
Find $(x,y)$ when $t=\pi$:
$(x,y)=(\sin2\pi+\cos \pi,\cos2\pi-\sin\pi)$
$(x,y)=(0+(-1),1-0)$
$(x,y)=(-1,1)$
Find the slope of the tangent to the curve at $(-1,1)$:
$m=\frac{dy}{dx}$
$m=\frac{dy/dt}{dx/dt}$
$m=\frac{-2\sin 2t-\cos t}{2\cos 2t-\sin t}|_{t=\pi}$
$m=\frac{-2\sin 2\pi -\cos \pi}{2\cos 2\pi-\sin\pi}$
$m=\frac{0-(-1)}{2-0}$
$m=\frac{1}{2}$
Find the equation of the tangent to the curve at $(-1,1)$:
$y-1=m(x-(-1))$
$y-1=\frac{1}{2}(x+1)$
$y-1=\frac{1}{2}x+\frac{1}{2}$
$y=\frac{1}{2}x+\frac{3}{2}$
