Animated Solution for Mathematics - Differentiation: Find all the tangents to the curve $y=\cos (x+y),-2\pi \le x\le 2\pi$, that are parallel to the line $x+2y=0$.

The given line is $x+2y=0$, which can be rewritten in slope-intercept form as $y=-\frac{1}{2}x$.

This reveals that the slope of the line is $m=-\frac{1}{2}$. Any line parallel to this must share the same slope, $m=-\frac{1}{2}$.

We are working with the implicit curve defined by $y=\cos (x+y)$. To find the slope of the tangent at any point, we differentiate both sides with respect to $x$, treating $y$ as a function of $x$.

Applying the chain rule to the right side, we obtain:

Expanding this expression, we get:

To isolate $\frac{dy}{dx}$, we move all terms containing the derivative to the left side:

Thus, the general expression for the slope of the tangent at any point $(x,y)$ on the curve is:

We set the derivative equal to our target slope of $-\frac{1}{2}$:

Cross-multiplying yields $2\sin (x+y)=1+\sin (x+y)$, which simplifies to the elegant condition:

Since $y=\cos (x+y)$, and we know that ${\sin }^2(\theta )+{\cos }^2(\theta )=1$, the condition $\sin (x+y)=1$ implies $\cos (x+y)=0$. Therefore, the $y$-coordinate of our points of tangency must be $y=0$.

Substituting $y=0$ into $\sin (x+y)=1$, we get $\sin (x)=1$. Within the interval $-2\pi \le x\le 2\pi$, this occurs at $x=\frac{\pi }{2}$ and $x=-\frac{3\pi }{2}$.

The points of tangency are $(\frac{\pi }{2},0)$ and $(-\frac{3\pi }{2},0)$. Using the point-slope form $y-y_1=m(x-x_1)$ with $m=-\frac{1}{2}$:

1. For $(\frac{\pi }{2},0)$: $y-0=-\frac{1}{2}(x-\frac{\pi }{2})\Rightarrow 2y=-x+\frac{\pi }{2}\Rightarrow$ $2x+4y-\pi =0$.

2. For $(-\frac{3\pi }{2},0)$: $y-0=-\frac{1}{2}(x+\frac{3\pi }{2})\Rightarrow 2y=-x-\frac{3\pi }{2}\Rightarrow$ $2x+4y+3\pi =0$.