Question

$16 \sin 12^\circ \cos 18^\circ \sin 48^\circ =$

• A. $\sqrt{10-2\sqrt{5}}$
• B. $\sqrt{10+2\sqrt{5}}$
• C. $\sqrt{5}-1$
• D. $\sqrt{5}+1$

Hint:

When evaluating products of sines and cosines, look for patterns that match product-to-sum formulas or the special product $\sin(60-A)\sin A \sin(60+A)$. Also, having the values of trig functions for angles like $18^\circ, 36^\circ, 54^\circ, 72^\circ$ memorized is extremely helpful. If your derivation is solid but doesn't match the answer key, consider the possibility of an error in the question paper.

Answer 1

The Correct Option is C

Step 1: Group Terms for Product-to-Sum: Expression: \( E = 16 \cos 18^\circ (\sin 48^\circ \sin 12^\circ) \). Use \( 2\sin A \sin B = \cos(A-B) - \cos(A+B) \). \[ 2 \sin 48^\circ \sin 12^\circ = \cos(36^\circ) - \cos(60^\circ) \] \[ = \frac{\sqrt{5}+1}{4} - \frac{1}{2} = \frac{\sqrt{5}+1-2}{4} = \frac{\sqrt{5}-1}{4} \] We know that \( \sin 18^\circ = \frac{\sqrt{5}-1}{4} \). So, \( 2 \sin 48^\circ \sin 12^\circ = \sin 18^\circ \).
Step 2: Substitute back into Expression: \[ E = 8 \cos 18^\circ (2 \sin 48^\circ \sin 12^\circ) \] \[ E = 8 \cos 18^\circ (\sin 18^\circ) \] \[ E = 4 (2 \sin 18^\circ \cos 18^\circ) \] \[ E = 4 \sin 36^\circ \]
Step 3: Evaluate \( 4 \sin 36^\circ \): We need to match the options. \( \sin 36^\circ = \sqrt{1 - \cos^2 36^\circ} = \sqrt{1 - \left(\frac{\sqrt{5}+1}{4}\right)^2} \) \[ \sin 36^\circ = \sqrt{1 - \frac{6+2\sqrt{5}}{16}} = \sqrt{\frac{16-6-2\sqrt{5}}{16}} = \frac{\sqrt{10-2\sqrt{5}}}{4} \] Therefore, \[ E = 4 \times \frac{\sqrt{10-2\sqrt{5}}}{4} = \sqrt{10-2\sqrt{5}} \]