Question

The value of \( 36 \big(4 \cos^2 9^\circ - 1\big) \big(4 \cos^2 27^\circ - 1\big) \big(4 \cos^2 81^\circ - 1\big) \big(4 \cos^2 243^\circ - 1\big) \) is:

• A. 18
• B. 27
• C. 36
• D. 54

Hint:

Use trigonometric identities like \( 4 \cos^2 \theta - 1 = \frac{\sin 3\theta}{\sin \theta} \) and simplify using periodicity of trigonometric functions.

Answer 1

The Correct Option is C

Using the trigonometric identity:

\[ 4 \cos^2 \theta - 1 = 4(1 - \sin^2 \theta) - 1 = 3 - 4 \sin^2 \theta = \frac{\sin 3\theta}{\sin \theta}. \]

Substitute this into the given expression:

\[ 36 \big(4 \cos^2 9^\circ - 1\big) \big(4 \cos^2 27^\circ - 1\big) \big(4 \cos^2 81^\circ - 1\big) \big(4 \cos^2 243^\circ - 1\big) \]

\[ = 36 \cdot \frac{\sin 27^\circ}{\sin 9^\circ} \cdot \frac{\sin 81^\circ}{\sin 27^\circ} \cdot \frac{\sin 243^\circ}{\sin 81^\circ} \cdot \frac{\sin 729^\circ}{\sin 243^\circ}. \]

Simplify the product:

\[ = 36 \cdot \frac{\sin 729^\circ}{\sin 9^\circ}. \]

Since \( \sin 729^\circ = \sin 9^\circ \) (as \( 729^\circ \mod 360^\circ = 9^\circ \)):

\[ \frac{\sin 729^\circ}{\sin 9^\circ} = 1. \]

Thus, the value is:

\[ 36 \cdot 1 = 36. \]