Let a and b be real numbers such that sina + sinb `=1/sqrt(2)`, cosa + cosb `=sqrt(3)/2`, then

Options

A. `cos(a-b)=1`
B. cos(a-b)=0
C. sin(a+b) =`sqrt(3)/2`
D. `sin(a-b)=1`

Answer

To solve the problem, we start with the given equations:

1. \( \sin a + \sin b = \frac{1}{\sqrt{2}} \)
2. \( \cos a + \cos b = \frac{\sqrt{3}}{2} \)

### Step 1: Use Sum-to-Product Identities

Using the sum-to-product identities, we can rewrite the sine and cosine sums:

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

\[ \cos a + \cos b = 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \]

### Step 2: Set Up the Equations

From the identities, we can set up the following equations:

\[ 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) = \frac{1}{\sqrt{2}} \quad (1) \]

\[ 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) = \frac{\sqrt{3}}{2} \quad (2) \]

### Step 3: Simplify the Equations

Dividing both equations by 2 gives:

\[ \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) = \frac{1}{2\sqrt{2}} \quad (3) \]

\[ \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) = \frac{\sqrt{3}}{4} \quad (4) \]

### Step 4: Divide Equation (3) by Equation (4)

Now, we can divide equation (3) by equation (4):

\[ \frac{\sin\left(\frac{a+b}{2}\right)}{\cos\left(\frac{a+b}{2}\right)} = \frac{\frac{1}{2\sqrt{2}}}{\frac{\sqrt{3}}{4}} \]

This simplifies to:

\[ \tan\left(\frac{a+b}{2}\right) = \frac{2}{\sqrt{6}} = \frac{1}{\sqrt{3}} \quad (5) \]

### Step 5: Find \( \frac{a+b}{2} \)

From equation (5), we know:

\[ \tan\left(\frac{a+b}{2}\right) = \frac{1}{\sqrt{3}} \]

This implies:

\[ \frac{a+b}{2} = \frac{\pi}{6} + n\pi \quad \text{for } n \in \mathbb{Z} \]

Thus,

\[ a+b = \frac{\pi}{3} + 2n\pi \quad (6) \]

### Step 6: Substitute Back to Find \( \cos\left(\frac{a-b}{2}\right) \)

Using equation (4):

\[ \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) = \frac{\sqrt{3}}{4} \]

We can find \( \cos\left(\frac{a+b}{2}\right) \):

\[ \cos\left(\frac{a+b}{2}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]

Substituting into equation (4):

\[ \frac{\sqrt{3}}{2} \cos\left(\frac{a-b}{2}\right) = \frac{\sqrt{3}}{4} \]

This simplifies to:

\[ \cos\left(\frac{a-b}{2}\right) = \frac{1}{2} \]

### Step 7: Find \( \frac{a-b}{2} \)

From \( \cos\left(\frac{a-b}{2}\right) = \frac{1}{2} \), we have:

\[ \frac{a-b}{2} = \frac{\pi}{3} + m\pi \quad \text{for } m \in \mathbb{Z} \]

Thus,

\[ a-b = \frac{2\pi}{3} + 2m\pi \quad (7) \]

### Step 8: Solve for \( a \) and \( b \)

Now we can solve the two equations (6) and (7):

1. \( a + b = \frac{\pi}{3} + 2n\pi \)
2. \( a - b = \frac{2\pi}{3} + 2m\pi \)

Adding these two equations:

\[ 2a = \frac{\pi}{3} + \frac{2\pi}{3} + 2(n+m)\pi \]

This gives:

\[ 2a = \pi + 2(n+m)\pi \implies a = \frac{\pi}{2} + (n+m)\pi \]

Subtracting the second equation from the first:

\[ 2b = \frac{\pi}{3} - \frac{2\pi}{3} + 2(n-m)\pi \]

This gives:

\[ 2b = -\frac{\pi}{3} + 2(n-m)\pi \implies b = -\frac{\pi}{6} + (n-m)\pi \]

### Conclusion

Thus, the values of \( a \) and \( b \) can be expressed in terms of integers \( n \) and \( m \).