What is the value of `sin292(1)/(2)`?

Options

A. `(1)/(3)sqrt(2+sqrt(3))`
B. `-(1)/(3)sqrt(2-sqrt(3))`
C. `(1)/(2)sqrt(2+sqrt(2))`
D. `-(1)/(2)sqrt(2+sqrt(2))`

Explanation

To find the value of \( \sin(292.5^\circ) \), we can follow these steps:

### Step 1: Rewrite the angle
We start by rewriting \( 292.5^\circ \) in a more manageable form. We can express it as:
\[ 292.5^\circ = \frac{585^\circ}{2} \]
This means we need to find \( \sin(292.5^\circ) = \sin\left(\frac{585^\circ}{2}\right) \).

### Step 2: Use the double angle formula
Using the double angle identity for sine, we have:
\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{2}} \]
In our case, \( \theta = 585^\circ \). Thus, we can write:
\[ \sin(292.5^\circ) = \sqrt{\frac{1 - \cos(585^\circ)}{2}} \]

### Step 3: Simplify \( \cos(585^\circ) \)
Next, we need to simplify \( \cos(585^\circ) \). Since \( 585^\circ \) is greater than \( 360^\circ \), we can reduce it:
\[ 585^\circ - 360^\circ = 225^\circ \]
Thus, \( \cos(585^\circ) = \cos(225^\circ) \).

### Step 4: Find \( \cos(225^\circ) \)
The angle \( 225^\circ \) is in the third quadrant, where cosine is negative. We know:
\[ \cos(225^\circ) = -\frac{1}{\sqrt{2}} \quad \text{(or } -\frac{\sqrt{2}}{2}\text{)} \]

### Step 5: Substitute back into the sine formula
Now we substitute \( \cos(585^\circ) \) back into our sine formula:
\[ \sin(292.5^\circ) = \sqrt{\frac{1 - (-\frac{1}{\sqrt{2}})}{2}} = \sqrt{\frac{1 + \frac{1}{\sqrt{2}}}{2}} \]

### Step 6: Simplify the expression
To simplify \( \frac{1 + \frac{1}{\sqrt{2}}}{2} \):
1. Convert \( 1 \) to a fraction with a common denominator:
\[ 1 = \frac{\sqrt{2}}{\sqrt{2}} \]
2. Now we have:
\[ \frac{\sqrt{2} + 1}{\sqrt{2}} \div 2 = \frac{\sqrt{2} + 1}{2\sqrt{2}} \]

### Step 7: Final simplification
Thus, we have:
\[ \sin(292.5^\circ) = \sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}} \]

### Step 8: Final answer
This can be further simplified, but it is already in a manageable form. The value of \( \sin(292.5^\circ) \) is:
\[ \sin(292.5^\circ) = \frac{\sqrt{\sqrt{2} + 1}}{\sqrt{2}} \]