For all pairs of angles (A, B), measured in degrees such that `sin A + sin B =sqrt(2)` and `cos A + cos B =sqrt(sqrt( 2))` , both hold simultaneously. The smallest possible value of `|A-B|` in degrees is

Options

A. (a) 15
B. (b) 30
C. (c) 45
D. (d) 60

Answer

To solve the problem, we need to find the smallest possible value of \(|A - B|\) given the equations:

1. \(\sin A + \sin B = \sqrt{2}\)
2. \(\cos A + \cos B = \sqrt[4]{2}\)

Let's break this down step by step:

### Step 1: Square both equations
We start by squaring both equations to eliminate the square roots.

- From equation 1:
\[ (\sin A + \sin B)^2 = (\sqrt{2})^2 \]
This expands to:
\[ \sin^2 A + \sin^2 B + 2\sin A \sin B = 2 \tag{Equation 3} \]

- From equation 2:
\[ (\cos A + \cos B)^2 = (\sqrt[4]{2})^2 \]
This expands to:
\[ \cos^2 A + \cos^2 B + 2\cos A \cos B = \sqrt{2} \tag{Equation 4} \]

### Step 2: Add the two equations
Now we add Equation 3 and Equation 4:
\[ (\sin^2 A + \sin^2 B + 2\sin A \sin B) + (\cos^2 A + \cos^2 B + 2\cos A \cos B) = 2 + \sqrt{2} \]

### Step 3: Use the Pythagorean identity
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can simplify:
\[ 1 + 1 + 2(\sin A \sin B + \cos A \cos B) = 2 + \sqrt{2} \]
This simplifies to:
\[ 2 + 2(\sin A \sin B + \cos A \cos B) = 2 + \sqrt{2} \]
Subtracting 2 from both sides gives:
\[ 2(\sin A \sin B + \cos A \cos B) = \sqrt{2} \]
Dividing by 2:
\[ \sin A \sin B + \cos A \cos B = \frac{\sqrt{2}}{2} \]

### Step 4: Apply the cosine of difference identity
We recognize that:
\[ \sin A \sin B + \cos A \cos B = \cos(A - B) \]
Thus, we have:
\[ \cos(A - B) = \frac{\sqrt{2}}{2} \]

### Step 5: Find the angles
The values of \(A - B\) that satisfy this equation are:
\[ A - B = 45^\circ + k \cdot 360^\circ \quad \text{or} \quad A - B = -45^\circ + k \cdot 360^\circ \quad \text{for any integer } k \]

### Step 6: Calculate the absolute difference
The smallest possible value of \(|A - B|\) is:
\[ |A - B| = 45^\circ \]

### Conclusion
Thus, the smallest possible value of \(|A - B|\) is \(45^\circ\).