If `asintheta+bcostheta=c`, find `acostheta-bsintheta=?`

Options

A. `pmsqrt(a^(2)+b^(2)+c^(2))`
B. `pmsqrt(a^(2)-b^(2)+c^(2))`
C. `pmsqrt(a^(2)+b^(2)-c^(2))`
D. none of these

Answer

To solve the problem where we need to find \( a \cos \theta - b \sin \theta \) given that \( a \sin \theta + b \cos \theta = c \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ a \sin \theta + b \cos \theta = c \]

### Step 2: Square both sides
Squaring both sides gives us:
\[ (a \sin \theta + b \cos \theta)^2 = c^2 \]

### Step 3: Expand the left side using the identity
Using the identity \( (x + y)^2 = x^2 + y^2 + 2xy \), we expand:
\[ a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta = c^2 \]

### Step 4: Define \( y \)
Let:
\[ y = a \cos \theta - b \sin \theta \]

### Step 5: Square \( y \)
Now, we square \( y \):
\[ (a \cos \theta - b \sin \theta)^2 = y^2 \]

### Step 6: Expand the squared expression
Using the identity \( (x - y)^2 = x^2 + y^2 - 2xy \), we expand:
\[ a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta = y^2 \]

### Step 7: Add the two equations
Now we add the two squared equations:
\[ (a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta) + (a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta) = c^2 + y^2 \]

### Step 8: Simplify the left side
Combining like terms:
\[ a^2 (\sin^2 \theta + \cos^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = c^2 + y^2 \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ a^2 + b^2 = c^2 + y^2 \]

### Step 9: Rearrange to find \( y^2 \)
Rearranging gives us:
\[ y^2 = a^2 + b^2 - c^2 \]

### Step 10: Solve for \( y \)
Taking the square root of both sides gives:
\[ y = \pm \sqrt{a^2 + b^2 - c^2} \]

### Conclusion
Thus, we find:
\[ a \cos \theta - b \sin \theta = \pm \sqrt{a^2 + b^2 - c^2} \]

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