Question

Find the equation of tangents to the curve $y=\cos (x+y),(–2\pi \le x\le 2\pi )$ that are parallel to the line $x+2y=0$.

Answer 1

$y=\cos (x+y)$
On differentiating $y$ w.r.t. $x$, we have
$\frac{dy}{dx}=\frac{-\sin (x+y)}{1+\sin (x+y)}$
$\therefore$ slope of tangent at $(x,y)$ $=\frac{-\sin (x+y)}{1+\sin (x+y)}$
Since the tangents to the curve are parallel to the line $x+2y=0$ , whose slope is $\frac{-1}{2}$, we have
$\frac{-\sin (x+y)}{1+\sin (x+y)}=\frac{-1}{2}$
$\sin (x+y)=1\Rightarrow x+y=n\pi +(-1{)}^n\frac{\pi }{2},n\in Z$
$\Rightarrow y=\cos (x+y)=\cos (n\pi +(-1{)}^n\frac{\pi }{2}),n\in Z\Rightarrow y=0\forall n\in Z$
Also, since $x\in [-2\pi ,2\pi ]$, we get $x=\frac{-3\pi }{2}$ and $x=\frac{\pi }{2}$.
Thus, tangents to the given curve are parallel to the line $x+2y=0$ only at points $(\frac{-3\pi }{2},0)$ and $(\frac{\pi }{2},0)$.
Therefore, the required equation of tangents are
$y-0=\frac{-1}{2}(x+\frac{3\pi }{2})\Rightarrow 2x+4y+3\pi =0$
and
$y-0=\frac{-1}{2}(x-\frac{\pi }{2})\Rightarrow 2x+4y-\pi =0$