Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Compute the number of ordered pairs $(x,y)$ that satisfy the system $$\begin{align*} \sin(x+y)&=\cos(x+y)\\ x^2+y^2&=\left(\dfrac{1995\pi}{4}\right)^2 \end{align*}$$

I got 5622, but I'm not sure if that's correct. Can someone provide a hint?

share|cite|improve this question
It would perhaps help if you gave some indication of your workings - which solutions have you found? Then it would be possible for us to identify whether you have missed something, or alternatively if you have double-counted somewhere. – Mark Bennet Apr 7 '12 at 21:46

Let $x=r\cos\theta$ and $y=r\sin\theta$ for $r=\frac{1995\pi}{4}$ parametrize the solutions on the circle given by the second equation. Then $\phi=x+y=r\sqrt2\sin\left(\theta+\frac\pi4\right)$ has equal sine and cosine iff $\phi\equiv\frac\pi4\pmod{\pi}$ i.e. $\iff\frac{\phi-\frac\pi4}{\pi}\in\mathbb{Z}$ $\iff\phi=\phi_k=\frac\pi4(4k+1)$ for $k\in\mathbb{Z}$. But $\left|\frac{\phi}{r\sqrt2}\right|\le1\iff$ $\left|4k+1\right|\le\frac{4r\sqrt2}{\pi}=1995\sqrt2\iff$ $\left|k+\frac14\right|\le\frac{r\sqrt2}{\pi}=\frac{1995\sqrt2}{4}\approx705.339$, $\iff k\in\{-705,\dots,705\}$, which entails $1411$ distinct, admisssible values of $\phi$.

To each one of these corresponds a line of slope $-1$ which intersects our circle at two points. So each $\phi_k\in(-1,1)$ furnishes two solutions $(x,y)$ & $(y,x)$ with $\theta=\theta_k=-\frac\pi4+\arcsin\frac{\phi_k}{r\sqrt2}$ and $(y,x)$ representing $\theta=\frac\pi2-\theta_k$, while if either of $\phi=\pm1$ were admissible (which isn't the case), it would only add one new solution. Thus we have $2\cdot1411=2822$ total solutions.

Thanks to @joriki for a second opinion on an earlier draft that stopped with $1411$.

share|cite|improve this answer
Yes, you're missing something. For each of those values of $k$, there are two different values of $\theta$, corresponding to $(x,y)$ and $(y,x)$, so there are $2\cdot1411=2822$ distinct solutions. Strange coincidence, since $5622=2\cdot2811$. – joriki Apr 7 '12 at 21:42

$x+ y = n \pi + \dfrac{\pi}{4}$ and $x^2 + y^2 = \left(\dfrac{1995 \pi}{4} \right)^2$. Letting $x = \pi X$ and $y = \pi Y$, we need number of solutions for $$X+Y =n + \dfrac14$$ and $$X^2 + Y^2 = \left(\dfrac{1995}4 \right)^2$$ Now $$(X+Y)^2 \leq 2(X^2 + Y^2)$$ Hence, $$\left( n + \dfrac14 \right)^2 \leq 2 \left(\dfrac{1995}4 \right)^2$$ $$-\dfrac{1995}4 \sqrt{2} \leq n + \dfrac14 \leq \dfrac{1995}4 \sqrt{2} \implies - \dfrac{1995 \sqrt{2} + 1}4 \leq n \leq \dfrac{1995 \sqrt{2} - 1}4$$ $$-705 \leq n\leq 705$$ For each $n$, we have two solution pairs $(x,y)$, since for each $n$ we have a quadratic in $X$ (or) $Y$.

Hence, the total number of solution pairs is $2 \times (705-(-705)+1) = 2 \times 1411 = 2822$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.