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

The equation of the tangent is $$y=-x+\frac{\pi}{2}+1$$ The equation of the normal is $$y=x+1-\frac{\pi}{2}$$ The graphs are shown below.
$$y=1+\cos x$$ 1) First, find the derivative of $y$: $$y'=0-\sin x=-\sin x$$ 2) Find the equation of the tangent at $(\pi/2,1)$: The slope of the tangent is $$y'(\pi/2)=-\sin\frac{\pi}{2}=-1$$ The equation of the tangent, therefore, is $$y-1=-1\Big(x-\frac{\pi}{2}\Big)$$ $$y-1=-x+\frac{\pi}{2}$$ $$y=-x+\frac{\pi}{2}+1$$ 3) Find the equation of the normal at $(\pi/2,1)$: Since the tangent and the normal at a point intersect at right angles, the product of their slopes equals $-1$. So if we call the slope of the normal at $(\pi/2,1)$ $s$, we have $$s\times(-1)=-1$$ $$s=1$$ The equation of the normal, therefore, is $$y-1=1\Big(x-\frac{\pi}{2}\Big)$$ $$y-1=x-\frac{\pi}{2}$$ $$y=x+1-\frac{\pi}{2}$$ The graphs of all the curves are lines are shown below.