Question

Evaluate: \[ (4\cos^2 9^\circ-3)(4\cos^2 27^\circ-3)= \]

• A. \(\sin 9^\circ\)
• B. \(\cos 9^\circ\)
• C. \(\tan 9^\circ\)
• D. \(\cot 9^\circ\)

Hint:

Use the identity \[ 4\cos^2 A-3=\frac{\cos 3A}{\cos A} \] when expressions contain terms like \(4\cos^2 A-3\).

Answer 1

The Correct Option is C

Step 1: Use the triple angle identity.
We know that \[ \cos 3A=4\cos^3 A-3\cos A \] Taking \(\cos A\) common, \[ \cos 3A=\cos A(4\cos^2 A-3) \] Therefore, \[ 4\cos^2 A-3=\frac{\cos 3A}{\cos A} \]

Step 2: Apply the identity to the first factor.
For \[ A=9^\circ, \] we get \[ 4\cos^2 9^\circ-3=\frac{\cos 27^\circ}{\cos 9^\circ} \]

Step 3: Apply the identity to the second factor.
For \[ A=27^\circ, \] we get \[ 4\cos^2 27^\circ-3=\frac{\cos 81^\circ}{\cos 27^\circ} \]

Step 4: Multiply both factors.
Now, \[ (4\cos^2 9^\circ-3)(4\cos^2 27^\circ-3) = \frac{\cos 27^\circ}{\cos 9^\circ} \cdot \frac{\cos 81^\circ}{\cos 27^\circ} \]

Step 5: Cancel the common factor.
Cancelling \[ \cos 27^\circ, \] we get \[ \frac{\cos 81^\circ}{\cos 9^\circ} \]

Step 6: Use complementary angle identity.
Since \[ 81^\circ=90^\circ-9^\circ, \] we have \[ \cos 81^\circ=\sin 9^\circ \] Therefore, \[ \frac{\cos 81^\circ}{\cos 9^\circ} = \frac{\sin 9^\circ}{\cos 9^\circ} = \tan 9^\circ \]

Step 7: Final conclusion.
Hence, \[ \boxed{\tan 9^\circ} \]