Question

\[ \cos 6^\circ \sin 24^\circ \cos 72^\circ= \]

• A. \(\frac{1}{4}\)
• B. \(-\frac{1}{8}\)
• C. \(-\frac{1}{4}\)
• D. \(\frac{1}{8}\)

Hint:

For products of trigonometric functions, use product-to-sum identities to simplify quickly.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are asked to find the product of three trigonometric values at specific angles.
Step 2: Key Formula or Approach:
1. Use product-to-sum identities: \( 2 \sin A \cos B = \sin(A+B) + \sin(A-B) \).
2. Recognize complementary relationships like \( \cos 72^\circ = \sin 18^\circ \).
Step 3: Detailed Explanation:

First, convert \( \cos 72^\circ \) to \( \sin 18^\circ \). The expression is \( \cos 6^\circ \sin 24^\circ \sin 18^\circ \).

Focus on \( \cos 6^\circ \sin 24^\circ \):
\[ \cos 6^\circ \sin 24^\circ = \frac{1}{2} [ \sin(24 + 6) + \sin(24 - 6) ] = \frac{1}{2} [ \sin 30^\circ + \sin 18^\circ ] \]

Substitute \( \sin 30^\circ = 1/2 \):
\[ \frac{1}{2} [ 1/2 + \sin 18^\circ ] = \frac{1}{4} + \frac{1}{2} \sin 18^\circ \]

Now multiply by the remaining factor \( \sin 18^\circ \):
\[ \left(\frac{1}{4} + \frac{1}{2} \sin 18^\circ\right) \sin 18^\circ = \frac{1}{4} \sin 18^\circ + \frac{1}{2} \sin^2 18^\circ \]

Recall \( \sin 18^\circ = \frac{\sqrt{5}-1}{4} \). Then \( \sin^2 18^\circ = \frac{5 + 1 - 2\sqrt{5}}{16} = \frac{6 - 2\sqrt{5}}{16} = \frac{3 - \sqrt{5}}{8} \).

Plug these in:
\[ \frac{1}{4} \left(\frac{\sqrt{5}-1}{4}\right) + \frac{1}{2} \left(\frac{3-\sqrt{5}}{8}\right) = \frac{\sqrt{5}-1}{16} + \frac{3-\sqrt{5}}{16} \]

Add the fractions:
\[ \frac{\sqrt{5} - 1 + 3 - \sqrt{5}}{16} = \frac{2}{16} = \frac{1}{8} \]

Step 4: Final Answer:
The value is 1/8.