What is the values of sin `15^(@)sin75^(@)`?

Options

A. `1//4`
B. `1//8`
C. `1//16`
D. 1

Explanation

To find the value of \( \sin 15^\circ \sin 75^\circ \), we can use the sine addition and subtraction formulas. Here’s a step-by-step solution:

### Step 1: Express \( \sin 15^\circ \) and \( \sin 75^\circ \) using known angles
We can express \( \sin 15^\circ \) and \( \sin 75^\circ \) as:
\[ \sin 15^\circ = \sin(45^\circ - 30^\circ) \]
\[ \sin 75^\circ = \sin(45^\circ + 30^\circ) \]

### Step 2: Use the sine subtraction and addition formulas
Using the sine subtraction formula:
\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]
For \( \sin 15^\circ \):
\[ \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \]
Substituting the known values:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \sin 30^\circ = \frac{1}{2} \]
So,
\[ \sin 15^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \]
\[ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

For \( \sin 75^\circ \):
Using the sine addition formula:
\[ \sin 75^\circ = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \]
Substituting the known values:
\[ \sin 75^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \]
\[ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

### Step 3: Multiply \( \sin 15^\circ \) and \( \sin 75^\circ \)
Now we can find \( \sin 15^\circ \sin 75^\circ \):
\[ \sin 15^\circ \sin 75^\circ = \left(\frac{\sqrt{3} - 1}{2\sqrt{2}}\right) \left(\frac{\sqrt{3} + 1}{2\sqrt{2}}\right) \]
Using the difference of squares:
\[ = \frac{(\sqrt{3})^2 - (1)^2}{(2\sqrt{2})^2} = \frac{3 - 1}{8} = \frac{2}{8} = \frac{1}{4} \]

### Final Answer
Thus, the value of \( \sin 15^\circ \sin 75^\circ \) is:
\[ \frac{1}{4} \]