1
$\begingroup$

The number of values of $(x,y)\epsilon (0,2\pi)$ for which $$\sin (x-y)+\cos(x-y)=0$$ My approach is as follows.

By observation we see that $x-y=3\pi/4, 7\pi/4...n\pi-\pi/4 $

Hence shouldn't there be infinite such $x$ and $y$ in the interval?

  • 0
    Yup, you are correct. – 2017-02-24
  • 0
    Thanks just wanted to confirm! – 2017-02-24

2 Answers 2

2

Yes

If $x,y \in [0,2\pi)$ then $x-2\pi \lt y \lt x+2\pi$

Meanwhile $\sin (x-y)+\cos(x-y) = 2\sin(\frac\pi 4)\sin(x-y+\frac14 \pi)$, which is zero when $\sin(x-y+\frac14 \pi)=0$

so the solutions are of the form of line segments:

  • $y=x+\frac54 \pi$ with $0 \le x \lt \frac34 \pi$ and $\frac54 \pi \le y \lt 2 \pi$
  • $y=x+\frac14 \pi$ with $0 \le x \lt \frac74 \pi$ and $\frac14 \pi \le y \lt 2 \pi$
  • $y=x-\frac34 \pi$ with $\frac34 \pi \le x \lt 2\pi$ and $0 \le y \lt \frac54 \pi$
  • $y=x-\frac74 \pi$ with $\frac74 \pi \le x \lt 2\pi$ and $0 \le y \lt \frac14 \pi$
3

Set $z=x-y$. Then the equation turns into $\tan z=-1$ which has solution $z=-\frac{\pi}{4}+k\pi$.

  • 0
    Hence an infinite x,y could be z in the interval right? – 2017-02-24
  • 1
    Absolutely. Choose $z=\frac{3}{4}\pi$ and fix, for example $x=0$, $y=\frac{3}{4}\pi$. Then for any $\epsilon>0$, the pair $(x+\epsilon,y+\epsilon)$ is also a solution ($\epsilon<\frac{5}{4}\pi$ of course). – 2017-02-24