`sin^(2) 75^(@) - sin^(2)15^(@)=`

Options

A. `1/2`
B. `sqrt(3)/2`
C. 1
D. 0

Answer

To solve the expression \( \sin^2 75^\circ - \sin^2 15^\circ \), we can use the identity for the difference of squares. The formula we will use is:

\[ A^2 - B^2 = (A + B)(A - B) \]

### Step-by-Step Solution:

1. **Identify A and B:**
Let \( A = \sin 75^\circ \) and \( B = \sin 15^\circ \).

2. **Apply the Difference of Squares Formula:**
Using the formula \( A^2 - B^2 = (A + B)(A - B) \), we have:
\[ \sin^2 75^\circ - \sin^2 15^\circ = (\sin 75^\circ + \sin 15^\circ)(\sin 75^\circ - \sin 15^\circ) \]

3. **Calculate \( \sin 75^\circ + \sin 15^\circ \):**
We can use the sine addition formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Here, \( A = 75^\circ \) and \( B = 15^\circ \):
\[ \sin 75^\circ + \sin 15^\circ = 2 \sin\left(\frac{75^\circ + 15^\circ}{2}\right) \cos\left(\frac{75^\circ - 15^\circ}{2}\right) \]
\[ = 2 \sin(45^\circ) \cos(30^\circ) \]

4. **Calculate \( \sin 75^\circ - \sin 15^\circ \):**
We can use the sine subtraction formula:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
\[ \sin 75^\circ - \sin 15^\circ = 2 \cos\left(\frac{75^\circ + 15^\circ}{2}\right) \sin\left(\frac{75^\circ - 15^\circ}{2}\right) \]
\[ = 2 \cos(45^\circ) \sin(30^\circ) \]

5. **Substituting Back:**
Now substituting back into our expression:
\[ \sin^2 75^\circ - \sin^2 15^\circ = (2 \sin(45^\circ) \cos(30^\circ))(2 \cos(45^\circ) \sin(30^\circ)) \]
\[ = 4 \sin(45^\circ) \cos(30^\circ) \cos(45^\circ) \sin(30^\circ) \]

6. **Substituting Values:**
Using known values:
- \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
- \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

We get:
\[ = 4 \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) \]
\[ = 4 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{4 \sqrt{3}}{8} = \frac{\sqrt{3}}{2} \]

### Final Answer:
\[ \sin^2 75^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{2} \]