The maximum value of `y=a cos x+b sin x` is

Options

A. `ab`
B. `1/(sqrt(a^(2)+b^(2)))`
C. `a^(2)+b^(2)`
D. `sqrt(a^(2)+b^(2))`

Explanation

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

### Step 1: Rewrite the expression
We know that \( y = a \cos x + b \sin x \). To find the maximum value, we can express this in a different form using the amplitude of the sine and cosine functions.

### Step 2: Find the amplitude
We can factor out the square root of the sum of the squares of \( a \) and \( b \):
\[ y = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \cos x + \frac{b}{\sqrt{a^2 + b^2}} \sin x \right) \]
Let \( R = \sqrt{a^2 + b^2} \).

### Step 3: Identify the coefficients
Define:
\[ \cos \alpha = \frac{a}{\sqrt{a^2 + b^2}} \quad \text{and} \quad \sin \alpha = \frac{b}{\sqrt{a^2 + b^2}} \]
This means we can rewrite \( y \) as:
\[ y = R (\cos \alpha \cos x + \sin \alpha \sin x) \]

### Step 4: Use the cosine addition formula
Using the cosine addition formula, we can simplify this to:
\[ y = R \cos(x - \alpha) \]

### Step 5: Find the maximum value
The maximum value of \( \cos(x - \alpha) \) is 1. Therefore, the maximum value of \( y \) is:
\[ y_{\text{max}} = R \cdot 1 = \sqrt{a^2 + b^2} \]

### Final Answer
Thus, the maximum value of \( y = a \cos x + b \sin x \) is:
\[ \boxed{\sqrt{a^2 + b^2}} \]