The greatest value of the function `f(x) = (sin 2x)/(sin(x + pi/4))` on the interval `[0,pi/2]` is

Options

A. `1/sqrt2`
B. `sqrt2`
C. `1`
D. `-sqrt2`

Answer

To find the greatest value of the function \( f(x) = \frac{\sin 2x}{\sin(x + \frac{\pi}{4})} \) on the interval \( [0, \frac{\pi}{2}] \), we can follow these steps:

### Step 1: Rewrite the Function
Using the double angle formula for sine, we can rewrite \( \sin 2x \):
\[ f(x) = \frac{2 \sin x \cos x}{\sin(x + \frac{\pi}{4})} \]

### Step 2: Expand the Denominator
Now, we expand \( \sin(x + \frac{\pi}{4}) \) using the sine addition formula:
\[ \sin(x + \frac{\pi}{4}) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}} = \frac{\sin x + \cos x}{\sqrt{2}} \]

### Step 3: Substitute Back into the Function
Substituting this back into our function gives:
\[ f(x) = \frac{2 \sin x \cos x}{\frac{\sin x + \cos x}{\sqrt{2}}} = \frac{2\sqrt{2} \sin x \cos x}{\sin x + \cos x} \]

### Step 4: Simplify the Function
Now, we simplify the function:
\[ f(x) = \frac{2\sqrt{2} \sin x \cos x}{\sin x + \cos x} \]

### Step 5: Analyze the Function
To find the maximum value of \( f(x) \), we can analyze the function. We know that \( \sin x \cos x \) achieves its maximum value of \( \frac{1}{2} \) at \( x = \frac{\pi}{4} \).

### Step 6: Evaluate at Critical Points
Evaluate \( f(x) \) at \( x = \frac{\pi}{4} \):
\[ f\left(\frac{\pi}{4}\right) = \frac{2\sqrt{2} \cdot \frac{1}{2}}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{1} = \sqrt{2} \]

### Step 7: Check Endpoints
Now, we check the endpoints of the interval:
- At \( x = 0 \):
\[ f(0) = \frac{\sin 0}{\sin(0 + \frac{\pi}{4})} = 0 \]
- At \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \frac{\sin \pi}{\sin(\frac{\pi}{2} + \frac{\pi}{4})} = 0 \]

### Conclusion
The maximum value of \( f(x) \) on the interval \( [0, \frac{\pi}{2}] \) occurs at \( x = \frac{\pi}{4} \):
\[ \text{Greatest value of } f(x) = \sqrt{2} \]