Question

find the equation of tangent to the curve$y$ =$\cos \left(x+y\right),-2\pi \le x\le 2\pi ,$ that is parallel to the line x + 2y = 0.

Answer 1

Given curve $y=\cos (x+y)$

We have find equation of tangent parallel to $x+2y=0$

slope of tangent is $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{d}{dx}\cos (x+y)=-\sin (x+y)(1+\frac{dy}{dx})$

$\Rightarrow \frac{dy}{dx}(1+\sin (x+y\left)\right)=-\sin (x+y)$

$\Rightarrow \frac{dy}{dx}=\frac{-\sin (x+y)}{1+\sin (x+y)}$

Now slope $x+2y=0$

is $1+2\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{-1}{2}$

Now +$\frac{\sin (x+y)}{1+\sin (x+y)}=+\frac{1}{2}$

$2\sin (x+y)=1+\sin (x+y)$

$\therefore y=\cos \left[\right(2n+1)\frac{x}{2}-y+y]=\cos \left[\right(2n+1\left)\frac{\pi }{2}\right]$

$\Rightarrow y=0$

Now $-2\pi \le x\le 2x$

Putting $n=0$ $x=\frac{\pi }{2}$

$n=-1$ $x=-\frac{3\pi }{2}$

Thus, $x=-\frac{3\pi }{2}$ & $x=\frac{\pi }{2}$

$\therefore$Points are $(-\frac{3\pi }{2},0)$ and $(\frac{\pi }{2},0)$

Now equation of tangent at $(-\frac{3\pi }{2},0)$ and $(\frac{\pi }{2},0)$ having

Solve $\frac{-1}{2}$ is

$2x+4y+3\pi =0$ and $2x+4y-\pi =0$