Question

$Smx+Smy=sin(x+y) \& |x|+|y|=1$

asfordered pireof $(x,y)=?$

$2sin(\frac{x+y}{2})col(\frac{x+y}{2})=2sin(\frac{x+y}{2})col(\frac{x+y}{2})$

$2sin(\frac{x+y}{2})[Col(\frac{x+y}{2})-Col(\frac{x+y}{2})]=0$

$2sin(\frac{x+y}{2})x2sin\frac{x}{2}sin\frac{y}{2}=0$

$sin(\frac{x+y}{2})=0$ or $sin\frac{x}{2}=0$ or $sin\frac{y}{2}=0$

$x+y=0$ or $x=0$ or $y=0$

Answer 1

6

Answer 2

The problem asks for the number of ordered pairs $(x,y)$ that satisfy two given equations:

1. $\sin x + \sin y = \sin(x+y)$
2. $|x| + |y| = 1$

Step 1: Simplify the first equation

We use the sum-to-product formula for $\sin x + \sin y$ and the double angle formula for $\sin(x+y)$.

$\sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$

$\sin(x+y) = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x+y}{2}\right)$

Substitute these into the first equation:

$2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x+y}{2}\right)$

Rearrange the terms:

$2 \sin\left(\frac{x+y}{2}\right) \left[ \cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right) \right] = 0$

Now, use the difference-to-product formula for cosines: $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

Let $A = \frac{x-y}{2}$ and $B = \frac{x+y}{2}$.

Then $\frac{A+B}{2} = \frac{\frac{x-y}{2} + \frac{x+y}{2}}{2} = \frac{x}{2}$.

And $\frac{A-B}{2} = \frac{\frac{x-y}{2} - \frac{x+y}{2}}{2} = \frac{-y}{2}$.

So, $\cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right) = -2 \sin\left(\frac{x}{2}\right) \sin\left(\frac{-y}{2}\right) = -2 \sin\left(\frac{x}{2}\right) (-\sin\left(\frac{y}{2}\right)) = 2 \sin\left(\frac{x}{2}\right) \sin\left(\frac{y}{2}\right)$.

Substitute this back into the equation:

$2 \sin\left(\frac{x+y}{2}\right) \left[ 2 \sin\left(\frac{x}{2}\right) \sin\left(\frac{y}{2}\right) \right] = 0$

$4 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x}{2}\right) \sin\left(\frac{y}{2}\right) = 0$

This equation holds if any of the factors are zero:

a) $\sin\left(\frac{x+y}{2}\right) = 0 \implies \frac{x+y}{2} = n\pi \implies x+y = 2n\pi$ for some integer $n$.

b) $\sin\left(\frac{x}{2}\right) = 0 \implies \frac{x}{2} = k\pi \implies x = 2k\pi$ for some integer $k$.

c) $\sin\left(\frac{y}{2}\right) = 0 \implies \frac{y}{2} = m\pi \implies y = 2m\pi$ for some integer $m$.

Step 2: Analyze the second equation

The second equation is $|x| + |y| = 1$. This equation represents a square with vertices at $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$.

From $|x| + |y| = 1$, we know that:

$-1 \le x \le 1$ and $-1 \le y \le 1$.

Now, we combine the conditions from Step 1 with this constraint.

Case 1: $x+y = 2n\pi$

Since $-1 \le x \le 1$ and $-1 \le y \le 1$, the sum $x+y$ must be in the range $[-2, 2]$.

So, $2n\pi$ must be in $[-2, 2]$.

For $n=0$, $x+y = 0$, which is in the range.

For $n=1$, $x+y = 2\pi \approx 6.28$, which is outside the range.

For $n=-1$, $x+y = -2\pi \approx -6.28$, which is outside the range.

Thus, the only possibility is $x+y=0$, or $y=-x$.

Substitute $y=-x$ into $|x|+|y|=1$:

$|x| + |-x| = 1$

$|x| + |x| = 1$

$2|x| = 1 \implies |x| = \frac{1}{2}$

This gives two possibilities for $x$:

• If $x = \frac{1}{2}$, then $y = -x = -\frac{1}{2}$. So, $(\frac{1}{2}, -\frac{1}{2})$ is a solution.

• If $x = -\frac{1}{2}$, then $y = -x = \frac{1}{2}$. So, $(-\frac{1}{2}, \frac{1}{2})$ is a solution.

Case 2: $x = 2k\pi$

Since $-1 \le x \le 1$, $2k\pi$ must be in $[-1, 1]$.

The only integer $k$ for which this is true is $k=0$.

So, $x=0$.

Substitute $x=0$ into $|x|+|y|=1$:

$|0| + |y| = 1 \implies |y| = 1$

This gives two possibilities for $y$:

• If $y=1$, then $x=0$. So, $(0,1)$ is a solution.

• If $y=-1$, then $x=0$. So, $(0,-1)$ is a solution.

Case 3: $y = 2m\pi$

Since $-1 \le y \le 1$, $2m\pi$ must be in $[-1, 1]$.

The only integer $m$ for which this is true is $m=0$.

So, $y=0$.

Substitute $y=0$ into $|x|+|y|=1$:

$|x| + |0| = 1 \implies |x| = 1$

This gives two possibilities for $x$:

• If $x=1$, then $y=0$. So, $(1,0)$ is a solution.

• If $x=-1$, then $y=0$. So, $(-1,0)$ is a solution.

Step 3: List all distinct solutions

The solutions found are:

1. $(\frac{1}{2}, -\frac{1}{2})$
2. $(-\frac{1}{2}, \frac{1}{2})$
3. $(0,1)$
4. $(0,-1)$
5. $(1,0)$
6. $(-1,0)$

All these 6 ordered pairs are distinct and satisfy both equations.

The problem asks for the number of ordered pairs $(x,y)$.

The final answer is $\boxed{6}$

Explanation of the solution:

The first equation $\sin x + \sin y = \sin(x+y)$ simplifies to $4 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x}{2}\right) \sin\left(\frac{y}{2}\right) = 0$. This implies that either $x+y=2n\pi$, $x=2k\pi$, or $y=2m\pi$ for integers $n, k, m$.

The second equation $|x|+|y|=1$ defines a square in the Cartesian plane with vertices at $(1,0), (0,1), (-1,0), (0,-1)$. This constraint limits the possible values of $x$ and $y$ to the range $[-1, 1]$.

Combining these conditions:

1. For $x+y=2n\pi$, the only integer $n$ that keeps $x+y$ within $[-2,2]$ is $n=0$, so $x+y=0$. Intersecting $y=-x$ with $|x|+|y|=1$ yields two points: $(\frac{1}{2}, -\frac{1}{2})$ and $(-\frac{1}{2}, \frac{1}{2})$.

2. For $x=2k\pi$, the only integer $k$ that keeps $x$ within $[-1,1]$ is $k=0$, so $x=0$. Intersecting $x=0$ (y-axis) with $|x|+|y|=1$ yields two points: $(0,1)$ and $(0,-1)$.

3. For $y=2m\pi$, the only integer $m$ that keeps $y$ within $[-1,1]$ is $m=0$, so $y=0$. Intersecting $y=0$ (x-axis) with $|x|+|y|=1$ yields two points: $(1,0)$ and $(-1,0)$.

All six points are distinct, hence there are 6 ordered pairs $(x,y)$ that satisfy both equations.