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: Understanding the Concept
This question involves simplifying a trigonometric product of sine and cosine functions. To solve it, we apply standard trigonometric identities such as product-to-sum and known trigonometric ratios of special angles like $18^\circ, 36^\circ,$ and $72^\circ$. The goal is to reduce the given product into a single trigonometric term that can be simplified further using known values.
Step 2: Key Formula or Approach
The key identities used are:
$2\sin A \cos B = \sin(A+B) + \sin(A-B)$
$2\sin A \sin B = \cos(A-B) - \cos(A+B)$
$\sin(90^\circ - x) = \cos x$
$\sin 18^\circ = \dfrac{\sqrt{5}-1}{4}$ and $\cos 36^\circ = \dfrac{\sqrt{5}+1}{4}$ These formulas will help in expressing the given product as a sum or difference of trigonometric functions, which can then be simplified using known standard values.
Step 3: Detailed Explanation
We start with: \[ E = 16 \sin 12^\circ \cos 18^\circ \sin 48^\circ \] Group and simplify: \[ E = 8\cos 18^\circ \times (2\sin 12^\circ \sin 48^\circ) \] Now, applying $2\sin A \sin B = \cos(A-B) - \cos(A+B)$: \[ E = 8\cos18^\circ[\cos(48^\circ - 12^\circ) - \cos(48^\circ + 12^\circ)] \] \[ E = 8\cos18^\circ(\cos36^\circ - \cos60^\circ) \] Substituting known values: \[ \cos 60^\circ = \frac{1}{2}, \quad \cos 36^\circ = \frac{\sqrt{5}+1}{4} \] Thus, \[ E = 8\cos18^\circ\left(\frac{\sqrt{5}+1}{4} - \frac{1}{2}\right) \] Instead of substituting $\cos18^\circ$ immediately, an alternate simplification is more efficient. We can use the trigonometric identity $\sin(60^\circ - A)\sin A\sin(60^\circ + A) = \frac{1}{4}\sin(3A)$, which relates a symmetric triple product of sines. Let $A = 12^\circ$, giving $\sin 12^\circ$, $\sin 48^\circ$, and $\sin 72^\circ = \sin(60^\circ + 12^\circ)$. Multiplying and dividing by $\sin 72^\circ$, we get: \[ E = 16 \cos 18^\circ \times \frac{\sin 12^\circ \sin 48^\circ \sin 72^\circ}{\sin 72^\circ} \] Using the above identity: \[ E = 16 \cos 18^\circ \times \frac{\frac{1}{4}\sin 36^\circ}{\sin 72^\circ} \] Since $\sin 72^\circ = 2\sin 36^\circ \cos 36^\circ$, we have: \[ E = 16 \cos 18^\circ \times \frac{\frac{1}{4}\sin 36^\circ}{2\sin 36^\circ \cos 36^\circ} = \frac{2\cos 18^\circ}{\cos 36^\circ} \] Now, use $\cos 18^\circ = \sin 72^\circ = 2\sin 36^\circ \cos 36^\circ$, giving: \[ E = \frac{2(2\sin 36^\circ \cos 36^\circ)}{\cos 36^\circ} = 4\sin 36^\circ \] Finally, substituting $\sin 36^\circ = \dfrac{\sqrt{10-2\sqrt{5}}}{4}$, we obtain: \[ E = 4 \times \frac{\sqrt{10-2\sqrt{5}}}{4} = \sqrt{10-2\sqrt{5}} \] Step 4: Final Answer
\[ \boxed{E = \sqrt{10 - 2\sqrt{5}}} \] Hence, the correct option is (A). The answer key’s option (C) $\sqrt{5}-1$ does not match this result, indicating an error in the provided key.