The number of ordered pairs (`alpha, beta`) ,where `alpha, beta in (-pi, pi)` satisfying `cos(alpha-beta) =1 "and" cos(alpha+beta) = (1)/(sqrt2)` is :

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to find the number of ordered pairs \((\alpha, \beta)\) where \(\alpha, \beta \in (-\pi, \pi)\) that satisfy the equations:

1. \(\cos(\alpha - \beta) = 1\)
2. \(\cos(\alpha + \beta) = \frac{1}{\sqrt{2}}\)

### Step 1: Solve \(\cos(\alpha - \beta) = 1\)

From the trigonometric identity, we know that \(\cos(x) = 1\) when \(x = 2n\pi\) for any integer \(n\). In our case, we have:

\[ \alpha - \beta = 2n\pi \]

Since \(\alpha\) and \(\beta\) are restricted to the interval \((- \pi, \pi)\), the only solution that fits this constraint is:

\[ \alpha - \beta = 0 \implies \alpha = \beta \]

### Step 2: Substitute \(\alpha = \beta\) into the second equation

Now we substitute \(\alpha = \beta\) into the second equation:

\[ \cos(\alpha + \beta) = \frac{1}{\sqrt{2}} \]

This simplifies to:

\[ \cos(2\alpha) = \frac{1}{\sqrt{2}} \]

### Step 3: Solve \(\cos(2\alpha) = \frac{1}{\sqrt{2}}\)

The cosine function equals \(\frac{1}{\sqrt{2}}\) at specific angles. Specifically:

\[ 2\alpha = \frac{\pi}{4} + 2k\pi \quad \text{and} \quad 2\alpha = -\frac{\pi}{4} + 2k\pi \]

for any integer \(k\).

### Step 4: Determine the values of \(\alpha\)

Dividing by 2 gives us:

\[ \alpha = \frac{\pi}{8} + k\pi \quad \text{and} \quad \alpha = -\frac{\pi}{8} + k\pi \]

### Step 5: Find valid \(k\) values

We need to find the values of \(\alpha\) that lie within the interval \((- \pi, \pi)\):

1. For \(k = 0\):
- \(\alpha = \frac{\pi}{8}\)
- \(\alpha = -\frac{\pi}{8}\)

2. For \(k = -1\):
- \(\alpha = \frac{\pi}{8} - \pi = -\frac{7\pi}{8}\)
- \(\alpha = -\frac{\pi}{8} - \pi = -\frac{9\pi}{8}\) (not valid since \(-\frac{9\pi}{8} < -\pi\))

3. For \(k = 1\):
- \(\alpha = \frac{\pi}{8} + \pi = \frac{9\pi}{8}\) (not valid since \(\frac{9\pi}{8} > \pi\))
- \(\alpha = -\frac{\pi}{8} + \pi = \frac{7\pi}{8}\)

### Step 6: Valid solutions for \(\alpha\)

The valid solutions for \(\alpha\) are:

1. \(\alpha = \frac{\pi}{8}\)
2. \(\alpha = -\frac{\pi}{8}\)
3. \(\alpha = -\frac{7\pi}{8}\)
4. \(\alpha = \frac{7\pi}{8}\)

### Step 7: Count ordered pairs \((\alpha, \beta)\)

Since \(\alpha = \beta\), each valid \(\alpha\) gives us a corresponding \(\beta\). Thus, we have:

- \((\frac{\pi}{8}, \frac{\pi}{8})\)
- \((- \frac{\pi}{8}, - \frac{\pi}{8})\)
- \((- \frac{7\pi}{8}, - \frac{7\pi}{8})\)
- \((\frac{7\pi}{8}, \frac{7\pi}{8})\)

This results in a total of **4 ordered pairs**.

### Final Answer

The number of ordered pairs \((\alpha, \beta)\) is **4**.

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