Question

\(\sin 15^\circ \sin 45^\circ \sin 75^\circ =\)

• A. \(\frac{\sqrt{2}}{4}\)
• B. \(\frac{\sqrt{2}}{2}\)
• C. \(\frac{\sqrt{2}}{8}\)
• D. \(-\frac{\sqrt{2}}{8}\)
• E. \(-\frac{\sqrt{2}}{4}\)

Hint:

Whenever you see angles like \(15^\circ\) and \(75^\circ\) in a product, always think of the identity \(\sin A \cos A = \frac{1}{2} \sin 2A\). It simplifies the calculation significantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We use trigonometric identities to simplify products of sines. Notice that \(75^\circ\) and \(15^\circ\) are complementary.

Step 2: Key Formula or Approach:
1. \(\sin(90^\circ - \theta) = \cos \theta\), so \(\sin 75^\circ = \cos 15^\circ\).
2. \(\sin \theta \cos \theta = \frac{1}{2} \sin(2\theta)\).
3. Value: \(\sin 45^\circ = \frac{1}{\sqrt{2}}\).

Step 3: Detailed Explanation:
The expression is:
\[ E = \sin 45^\circ \cdot (\sin 15^\circ \cdot \sin 75^\circ) \]
Substitute \(\sin 75^\circ\) with \(\cos 15^\circ\):
\[ E = \frac{1}{\sqrt{2}} \cdot (\sin 15^\circ \cdot \cos 15^\circ) \]
Apply the double angle identity:
\[ E = \frac{1}{\sqrt{2}} \cdot \left( \frac{1}{2} \sin(2 \times 15^\circ) \right) \]
\[ E = \frac{1}{2\sqrt{2}} \cdot \sin 30^\circ \]
Substitute \(\sin 30^\circ = \frac{1}{2}\):
\[ E = \frac{1}{2\sqrt{2}} \cdot \frac{1}{2} = \frac{1}{4\sqrt{2}} \]
Rationalize the denominator by multiplying by \(\frac{\sqrt{2}}{\sqrt{2}}\):
\[ E = \frac{\sqrt{2}}{4 \times 2} = \frac{\sqrt{2}}{8} \]

Step 4: Final Answer:
The value is \(\frac{\sqrt{2}}{8}\).