cos(x)-cos(y)=a; sin(x)-sin(y)=b; tìm x,y

To solve the system of trigonometric equations:

\[
\cos(x) - \cos(y) = a \quad \text{(1)}
\]
\[
\sin(x) - \sin(y) = b \quad \text{(2)}
\]

we can use trigonometric identities and manipulation.

### Step 1: Use sum-to-product identities

We know the following sum-to-product identities:

\[
\cos(A) - \cos(B) = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)
\]
\[
\sin(A) - \sin(B) = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)
\]

### Step 2: Apply these identities to the given equations

For equation (1):
\[
\cos(x) - \cos(y) = -2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = a
\]
For equation (2):
\[
\sin(x) - \sin(y) = 2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = b
\]

Now we have the system:
\[
-2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = a
\]
\[
2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = b
\]

### Step 3: Solve for \(\sin\left(\frac{x-y}{2}\right)\)

Divide the two equations:
\[
\frac{-2 \sin\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)}{2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right)} = \frac{a}{b}
\]
Simplifying:
\[
\frac{- \sin\left(\frac{x+y}{2}\right)}{\cos\left(\frac{x+y}{2}\right)} = \frac{a}{b}
\]
This gives:
\[
- \tan\left(\frac{x+y}{2}\right) = \frac{a}{b}
\]
So:
\[
\tan\left(\frac{x+y}{2}\right) = -\frac{a}{b}
\]

### Step 4: Solve for \(\frac{x+y}{2}\)

Thus:
\[
\frac{x+y}{2} = \tan^{-1}\left(-\frac{a}{b}\right)
\]
So:
\[
x + y = 2 \tan^{-1}\left(-\frac{a}{b}\right)
\]

### Step 5: Solve for \(\sin\left(\frac{x-y}{2}\right)\)

Substitute \(\tan\left(\frac{x+y}{2}\right)\) back into one of the original equations (for simplicity, the second one):
\[
2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) = b
\]
Since we know \(\frac{x+y}{2}\), we can solve for \(\sin\left(\frac{x-y}{2}\right)\).

Finally, after solving \(\sin\left(\frac{x-y}{2}\right)\) and \(\frac{x+y}{2}\), we can find both \(x\) and \(y\) by solving the system:
\[
x = \frac{x+y}{2} + \frac{x-y}{2}
\]
\[
y = \frac{x+y}{2} - \frac{x-y}{2}
\]

Would you like me to continue with detailed calculations or clarifications?

### Further Questions:
1. What are the possible solutions for \(x\) and \(y\) in different quadrants?
2. How do we handle cases where \(\tan^{-1}(-a/b)\) gives multiple solutions?
3. Can these equations have a geometric interpretation on the unit circle?
4. How do the signs of \(a\) and \(b\) affect the solutions?
5. What happens if \(a = 0\) or \(b = 0\)?

**Tip:** When working with trigonometric identities, sum-to-product formulas are often useful in simplifying differences between functions.

cos(x) - cos(y) = a; sin(x) - sin(y) = b; tìm x,y

Discover how to solve the trigonometric system cos(x) - cos(y) = a and sin(x) - sin(y) = b using sum-to-product identities and inverse trigonometric functions. This solution is suitable for advanced high school and college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation