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

Options

A. `1/2`
B. `sqrt3/2`
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, which states that \( a^2 - b^2 = (a + b)(a - b) \).

### Step-by-Step Solution:

1. **Identify the expression**:
\[ \sin^2 75^\circ - \sin^2 15^\circ \]

2. **Apply the difference of squares formula**:
\[ \sin^2 75^\circ - \sin^2 15^\circ = (\sin 75^\circ + \sin 15^\circ)(\sin 75^\circ - \sin 15^\circ) \]

3. **Use the sine addition formula**:
The formula for \( \sin A + \sin B \) is:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Here, let \( A = 75^\circ \) and \( B = 15^\circ \):
\[ \sin 75^\circ + \sin 15^\circ = 2 \sin\left(\frac{75^\circ + 15^\circ}{2}\right) \cos\left(\frac{75^\circ - 15^\circ}{2}\right) \]

4. **Calculate the angles**:
\[ \frac{75^\circ + 15^\circ}{2} = \frac{90^\circ}{2} = 45^\circ \]
\[ \frac{75^\circ - 15^\circ}{2} = \frac{60^\circ}{2} = 30^\circ \]

5. **Substitute back into the equation**:
\[ \sin 75^\circ + \sin 15^\circ = 2 \sin(45^\circ) \cos(30^\circ) \]

6. **Use known values**:
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}}, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
Therefore,
\[ \sin 75^\circ + \sin 15^\circ = 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} \]

7. **Use the sine subtraction formula**:
The formula for \( \sin A - \sin B \) is:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
Using the same angles:
\[ \sin 75^\circ - \sin 15^\circ = 2 \cos(45^\circ) \sin(30^\circ) \]

8. **Substitute known values**:
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}}, \quad \sin(30^\circ) = \frac{1}{2} \]
Therefore,
\[ \sin 75^\circ - \sin 15^\circ = 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{1}{\sqrt{2}} \]

9. **Combine the results**:
Now substituting back into the difference of squares:
\[ \sin^2 75^\circ - \sin^2 15^\circ = \left(\frac{\sqrt{6}}{2}\right) \left(\frac{1}{\sqrt{2}}\right) \]

10. **Final calculation**:
\[ = \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{6}}{\sqrt{4}} = \frac{\sqrt{6}}{2} \]

### Final Answer:
\[ \sin^2 75^\circ - \sin^2 15^\circ = \frac{\sqrt{6}}{2} \]