## University Calculus: Early Transcendentals (3rd Edition)

$$y=f(x)=\sin x\hspace{1cm}\{-3\pi/2\le x\le 2\pi\}$$ The derivative of $f(x)$: $$f'(x)=\cos x$$ The tangent lines to $f(x)$ are: 1) At $x=-\pi: f(x)=\sin(-\pi)=0$ and $f'(x)=\cos(-\pi)=-1$ $$(y-0)=-1(x-(-\pi))$$ $$y=-(x+\pi)=-x-\pi$$ 2) At $x=0: f(x)=\sin(0)=0$ and $f'(x)=\cos(0)=1$ $$(y-0)=1(x-0)$$ $$y=x$$ 3) At $x=3\pi/2: f(x)=\sin(3\pi/2)=-1$ and $f'(x)=\cos(3\pi/2)=0$ $$(y-(-1))=0(x-\frac{3\pi}{2})$$ $$y+1=0$$ $$y=-1$$ The graphs of the curve and its tangent lines are below.