If a cos `theta` + b sin `theta`= c, then what is the value of a sin `theta`- b cos `theta` ?

Answer

To solve the problem, we need to find the value of \( a \sin \theta - b \cos \theta \) given that \( a \cos \theta + b \sin \theta = c \).

### Step-by-Step Solution:

1. **Define the Expression**:
Let \( x = a \sin \theta - b \cos \theta \). Our goal is to find the value of \( x \).

2. **Square Both Sides**:
We square \( x \):
\[ x^2 = (a \sin \theta - b \cos \theta)^2 \]

3. **Expand the Squared Expression**:
Using the identity \( (p - q)^2 = p^2 - 2pq + q^2 \):
\[ x^2 = a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta \]

4. **Square the Given Equation**:
Now, square the equation \( a \cos \theta + b \sin \theta = c \):
\[ c^2 = (a \cos \theta + b \sin \theta)^2 \]

5. **Expand This Expression**:
Again using the identity:
\[ c^2 = a^2 \cos^2 \theta + 2ab \sin \theta \cos \theta + b^2 \sin^2 \theta \]

6. **Combine the Two Equations**:
Now we have:
\[ x^2 = a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta \]
\[ c^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta \]

7. **Add the Two Equations**:
Adding the two equations:
\[ x^2 + c^2 = (a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta) + (a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta) \]

8. **Simplify the Combined Equation**:
The terms \( -2ab \sin \theta \cos \theta \) and \( +2ab \sin \theta \cos \theta \) cancel out:
\[ x^2 + c^2 = a^2 (\sin^2 \theta + \cos^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) \]

9. **Utilize the Pythagorean Identity**:
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ x^2 + c^2 = a^2 + b^2 \]

10. **Rearranging for \( x^2 \)**:
\[ x^2 = a^2 + b^2 - c^2 \]

11. **Taking the Square Root**:
To find \( x \):
\[ x = \sqrt{a^2 + b^2 - c^2} \]

12. **Final Result**:
Thus, the value of \( a \sin \theta - b \cos \theta \) is:
\[ a \sin \theta - b \cos \theta = \sqrt{a^2 + b^2 - c^2} \]