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:
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:
Show Hint
- 18
- 27
- 36
- 54
The Correct Option is C
Solution and Explanation
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:
\[ 36 \cdot 1 = 36. \]
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