`sin (A - 45 ^(@) ) = (1)/( sqrt2) (sin A - cos A) `

Answer

To solve the equation \( \sin(A - 45^\circ) = \frac{1}{\sqrt{2}}(\sin A - \cos A) \), we will start by proving that the left-hand side (LHS) is equal to the right-hand side (RHS).

### Step 1: Write down the equation
We have:
\[ \sin(A - 45^\circ) = \frac{1}{\sqrt{2}}(\sin A - \cos A) \]

### Step 2: Use the sine difference formula
Recall the sine difference formula:
\[ \sin(x - y) = \sin x \cos y - \cos x \sin y \]
In our case, let \( x = A \) and \( y = 45^\circ \). Therefore:
\[ \sin(A - 45^\circ) = \sin A \cos 45^\circ - \cos A \sin 45^\circ \]

### Step 3: Substitute the values of \( \sin 45^\circ \) and \( \cos 45^\circ \)
We know that:
\[ \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \]
Substituting these values into the equation gives:
\[ \sin(A - 45^\circ) = \sin A \cdot \frac{1}{\sqrt{2}} - \cos A \cdot \frac{1}{\sqrt{2}} \]

### Step 4: Factor out \( \frac{1}{\sqrt{2}} \)
This simplifies to:
\[ \sin(A - 45^\circ) = \frac{1}{\sqrt{2}}(\sin A - \cos A) \]

### Step 5: Compare LHS and RHS
Now we can see that:
\[ \sin(A - 45^\circ) = \frac{1}{\sqrt{2}}(\sin A - \cos A) \]
This shows that the left-hand side is equal to the right-hand side.

### Conclusion
Thus, we have proved that:
\[ \sin(A - 45^\circ) = \frac{1}{\sqrt{2}}(\sin A - \cos A) \]