Question

Find the equation of tangent of the curve $y = \cos(x+y)$, $-2\pi \leq x \leq 2\pi$, which is parallel to the line $x + 2y = 0$.

Hint:

For tangents parallel to a line: find slope from given line, equate with $\dfrac{dy}{dx}$ of curve, then find points and equations.

Answer 1

Step 1: Slope of the given line. Equation: $x + 2y = 0 \;\Rightarrow\; y = -\tfrac{1}{2}x$. So, slope = $-\tfrac{1}{2}$.

Step 2: Differentiate the curve. Curve: \[ y = \cos(x+y) \] Differentiate both sides w.r.t. $x$: \[ \frac{dy}{dx} = -\sin(x+y)\left(1 + \frac{dy}{dx}\right) \] \[ \frac{dy}{dx} + \sin(x+y)\frac{dy}{dx} = -\sin(x+y) \] \[ \frac{dy}{dx}(1 + \sin(x+y)) = -\sin(x+y) \] \[ \frac{dy}{dx} = \frac{-\sin(x+y)}{1 + \sin(x+y)} \]

Step 3: Condition for parallelism. We need $\dfrac{dy}{dx} = -\tfrac{1}{2}$. \[ \frac{-\sin(x+y)}{1 + \sin(x+y)} = -\frac{1}{2} \] \[ \frac{\sin(x+y)}{1 + \sin(x+y)} = \frac{1}{2} \] \[ 2 \sin(x+y) = 1 + \sin(x+y) $\Rightarrow$ \sin(x+y) = 1 \]

Step 4: Corresponding points. If $\sin(x+y) = 1 \;\Rightarrow\; x+y = \frac{\pi}{2} + 2n\pi$. Also, from curve $y = \cos(x+y) = \cos\left(\frac{\pi}{2} + 2n\pi\right) = 0$. So, $y=0$, $x = \frac{\pi}{2} + 2n\pi$. For $-2\pi \leq x \leq 2\pi$, possible $x = -\tfrac{3\pi}{2}, \; \tfrac{\pi}{2}$.

Step 5: Equation of tangent. Equation of tangent: $y - y_{1} = m(x - x_{1})$, with $m = -\tfrac{1}{2}$. At $\left(\tfrac{\pi}{2},0\right)$: \[ y - 0 = -\tfrac{1}{2}(x - \tfrac{\pi}{2}) \] \[ x + 2y - \tfrac{\pi}{2} = 0 \] At $\left(-\tfrac{3\pi}{2},0\right)$: \[ y - 0 = -\tfrac{1}{2}(x + \tfrac{3\pi}{2}) \] \[ x + 2y + \tfrac{3\pi}{2} = 0 \]

Final Answer: \[ \boxed{\; x + 2y - \tfrac{\pi}{2} = 0 \text{or} x + 2y + \tfrac{3\pi}{2} = 0 \;} \]