Find coordinates of a point Q on the graph $\sin (x) + \cos (y) = 0.5$ given that the gradient of its tangent is perpendicular to point P.

Question

Note:
Point $P$ is on the $y$-axis and above the $x$-axis

$\frac{-\pi}{6}\le x \le\frac{7\pi}{6}$

$\frac{-2\pi}{3}\le y\le\frac{2\pi}{3}$

What I have done so far:
Solving for $P$:
$$x = 0 \\ \sin (0) + \cos (y) = 0.5 \\ 0 + \cos (y) = 0.5 \\ y= \pm\frac{\pi}{3} $$

For $P$, $y \gt 0$
$\therefore y = \frac{\pi}{3}$
Solving for $\frac{dy}{dx}$:
$$\sin(x) + \cos(y) = 0.5 \\ \cos(x) - \sin(y)\frac{dy}{dx} = 0$$ $\therefore \frac{dy}{dx} = \frac{\cos(x)}{\sin (y)}$
Derivative at $P$:
$$\frac{dy}{dx} = \frac{\cos(x)}{\sin (y)} = \frac{\cos(0)}{\sin(\frac{\pi}{3})} = \frac{2}{\sqrt3}$$ As for the gradient of the tangent line at $Q$ is perpendicular to that at $P$: $\frac{dy}{dx} = \frac{-\sqrt3}{2}$
How do I solve for the coordinates of $Q$ after this?

Answer 1

Good work. Note the point $Q$ lies on the curve as well as the tangent line. So: $$\begin{cases}\frac{\cos x}{\sin y}=-\frac{\sqrt{3}}{2}\\ \sin x+\cos y=0.5 \end{cases} \Rightarrow$$ From the first: $$\cos^2x=\frac34(1-\cos^2y)\Rightarrow \cos y=\pm\sqrt{1-\frac43\cos^2x}$$ Now sub it to the second: $$1-\frac43\cos^2x=\frac14-\sin x+\sin^2x\Rightarrow \\ 4\sin^2x+12\sin x-7=0 \Rightarrow \sin x=\frac12.$$ Referring to the given constraints, the final answer is: $$x=\pi-\frac{\pi}{6},y=\frac{\pi}{2}$$

Answer 2

Note that the curve (one of an infinite number of such "loops" satisfying the given equation) is symmetrical about the point $ \ \left(\frac{\pi}{2} \ , \ 0 \right) \ ; $ this is a consequence of the trigonometric function properties $ \ \cos (-y) \ = \ \cos y \ $ and $ \ \sin \left(\frac{\pi}{2} + \xi \right) \ = \ \sin \left(\frac{\pi}{2} - \xi \right) \ \ . $ If we use this latter symmetry to define a coordinate $ \ x \ = \ \frac{\pi}{2} + \xi \ \Rightarrow \ \xi \ = \ x - \frac{\pi}{2} \ \ , $ then the curve equation becomes $ \ \sin \left(\frac{\pi}{2} + \xi \right) + \cos y \ = \ 0.5 \ $ over the intervals $ \ \frac{-2\pi}{3} \ \le \ \xi \ , \ y \ \le \ \frac{2\pi}{3} \ \ . $ The slope of the tangent line at $ \ P(x,y) \ $ is then

$$ \frac{dy}{dx}|_P \ \ = \ \ \frac{dy}{d \xi}|_P \ \ = \ \ \frac{\cos\left(\frac{\pi}{2} + \xi_P \right)}{\sin (y_P)} $$

and the slope of the normal line (which is equal to the slope of the tangent line at $ \ Q \ $ ) is

$$ \frac{dy}{dx}|_Q \ \ = \ \ -\frac{\sin (y_P)}{\cos\left(\frac{\pi}{2} + \xi_P \right)} \ \ = \ \ \frac{\cos\left(\frac{\pi}{2} + \xi_Q \right)}{\sin (y_Q)} \ \ , $$

The "angle-addition" formula for cosine yields $ \ \cos\left(\frac{\pi}{2} + \xi_Q \right) \ = \ \cos\left(\frac{\pi}{2} \right)·\cos \xi_Q \ - \ \sin\left(\frac{\pi}{2}\right) ·\sin \xi_Q $ $ = \ -\sin \xi_Q \ \ , $ and similarly, $ \ \cos\left(\frac{\pi}{2} + y_Q \right) \ = \ -\sin y_Q \ \ . $ Thus, we may write $$ \frac{dy}{dx}|_Q \ \ = \ \ \frac{-\sin \xi_Q}{-\cos\left(\frac{\pi}{2} + y_Q \right)} \ \ = \ \ \frac{\sin \xi_Q}{\cos\left(\frac{\pi}{2} + y_Q \right)} \ \ = \ \ -\frac{\sin (y_P)}{\cos\left(\frac{\pi}{2} + \xi_P \right)} \ \ , $$ which now suggests how to make correspondences between coordinates.

You identified $ \ P \ $ as one of the $ \ y-$intercepts $ \ (x_P \ = \ 0 \Rightarrow \ \xi_P \ = \ -\frac{\pi}{2} ) \ $ of the curve, which has $ \ \cos(y_P) \ = \ 0.5 \ \Rightarrow \ y_P \ = \ \frac{\pi}{3} \ \ . $ We may then take $$ \sin(y_P) \ \ = \ \ \frac{\sqrt3}{2} \ \ = \ \ \sin(\xi_Q) \ \ \ , \ \ \ \cos\left(\frac{\pi}{2} + \xi_P \right) \ \ = \ \ 1 \ \ = \ \ -\cos\left(\frac{\pi}{2} + y_Q \right) \ \ , $$ or $$ \sin(y_P) \ \ = \ \ \frac{\sqrt3}{2} \ \ = \ \ -\sin(\xi_Q) \ \ \ , \ \ \ \cos\left(\frac{\pi}{2} + \xi_P \right) \ \ = \ \ 1 \ \ = \ \ \cos\left(\frac{\pi}{2} + y_Q \right) \ \ . $$

The first pair of equations gives us $ \ \sin(\xi_Q) \ = \ \frac{\sqrt3}{2} \ \Rightarrow \ \xi_Q \ = \ \frac{\pi}{3} \ , \ \frac{2\pi}{3} \ \Rightarrow \ x_Q \ = \ \frac{5\pi}{6} \ , \ \frac{7\pi}{6} \ $ and $ \ \cos\left(\frac{\pi}{2} + y_Q \right) \ = \ -1 \ \Rightarrow \ \frac{\pi}{2} + y_Q \ = \ \pi \ \Rightarrow \ y_Q \ = \ \frac{\pi}{2} \ \ ; $ of these two points, $ \ \sin \left( \frac{5\pi}{6} \right) \ + \ \cos \left( \frac{\pi}{2} \right) \ = \ 0.5 \ $ satisfies the curve equation. On the other hand, the second pair produces $ \ \sin(\xi_Q) \ = \ -\frac{\sqrt3}{2} $ $ \Rightarrow \ \xi_Q \ = \ -\frac{\pi}{3} \ , \ -\frac{2\pi}{3} \ \Rightarrow \ x_Q \ = \ \frac{\pi}{6} \ , \ -\frac{\pi}{6} \ $ and $ \ \cos\left(\frac{\pi}{2} + y_Q \right) \ = \ 1 \ \Rightarrow \ \frac{\pi}{2} + y_Q \ = \ 0 \ \Rightarrow \ y_Q \ = \ -\frac{\pi}{2} \ \ , $ for which only $ \ \sin \left( \frac{\pi}{6} \right) \ + \ \cos \left( -\frac{\pi}{2} \right) \ = \ 0.5 \ $ works in the equation.

This tells us that the location of point $ \ Q \ $ is $ \ \left(\frac{5\pi}{6} \ , \ \frac{\pi}{2} \right) \ \ . $ We have also found, as we would expect from the symmetry of this loop about $ \ \left(\frac{\pi}{2} \ , \ 0 \right) \ , $ that there is a second point $ \ Q' \ \left(\frac{\pi}{6} \ , \ \frac{-\pi}{2} \right) \ $ at which the tangent line is perpendicular to the tangent line at $ \ P \ \ . $

By a related argument, we can also show that there is a point $ \ P' \ \left( \pi \ , \ \frac{-\pi}{3} \right) \ $ with a tangent line parallel to the one at $ \ P \ \ . $ Naturally, because the full "curve" described by the equation covers the plane with "loops" with periodicity $ \ 2 \pi \ $ in the $ \ x-$ and $ \ y-$directions, there are "families" of points $ \ \mathcal{P} \ \left(2 m \pi \ , \ \frac{\pi}{3} + 2 n \pi \right) \ \ $ and $ \ \mathcal{Q} \ \left(\frac{5\pi}{6} + 2 m \pi \ , \ \frac{\pi}{2} + 2 n \pi \right) \ \ , \ m \ \ \text{and} \ \ n \ $ being integers, which have the prescribed relationship.