The Maximum value of `a sin theta+b cos theta` is:

Options

A. `sqrt(a^(2)-b^(2))`
B. `sqrt(a^(2)+b^(2))`
C. `-sqrt(a^(2)+b^(2))`
D. `sqrt(b^(2)-a^(2))`

Answer

To find the maximum value of the expression \( a \sin \theta + b \cos \theta \), we can use the following steps:

### Step 1: Identify the Expression
We start with the expression:
\[ f(\theta) = a \sin \theta + b \cos \theta \]

### Step 2: Use the Formula for Maximum Value
The maximum value of the expression \( a \sin \theta + b \cos \theta \) can be derived using the formula:
\[ \text{Maximum value} = \sqrt{a^2 + b^2} \]

### Step 3: Justification of the Formula
This formula comes from the fact that \( a \sin \theta + b \cos \theta \) can be rewritten in the form of a single sinusoidal function. Specifically, we can express it as:
\[ R \sin(\theta + \phi) \]
where \( R = \sqrt{a^2 + b^2} \) and \( \tan \phi = \frac{b}{a} \). The maximum value of \( R \sin(\theta + \phi) \) is \( R \), which occurs when \( \sin(\theta + \phi) = 1 \).

### Step 4: Conclusion
Thus, the maximum value of \( a \sin \theta + b \cos \theta \) is:
\[ \sqrt{a^2 + b^2} \]

### Final Answer
The maximum value of \( a \sin \theta + b \cos \theta \) is \( \sqrt{a^2 + b^2} \).

---