Question

$\sin {12}^o\sin {48}^o\sin {54}^o$ is equal to

• A. $\frac{1}{16}$
• B. $\frac{1}{32}$
• C. $\frac{1}{8}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C

$\frac{1}{8}$

Explanation for the correct answer:

Simplifying the given expression:

$$\begin{gathered} \Rightarrow \sin 12°\sin 48°\sin 54° \\ \Rightarrow \frac{1}{2}(2\sin 12°\sin 48°)\sin 54° \end{gathered}$$

Using the formula $2\sin x\sin y=\cos (x–y)–\cos (x+y)$

$$\begin{gathered} \Rightarrow \frac{1}{2}\left[\cos (12°–48°)–\cos (12°+48°)\right]\sin 54° \\ \Rightarrow \frac{1}{2}\left[\cos 36°–\cos 60°\right]\sin 54° \\ \Rightarrow \frac{1}{2}\left[\cos 36°\sin 54°–\left(\frac{1}{2}\right)\sin 54°\right] \\ \Rightarrow \frac{1}{4}\left[2\cos 36°\sin 54°–\sin 54°\right] \end{gathered}$$]

Using the formula $2\cos x\sin y=\sin (x+y)–\sin (x–y)$

$$\begin{gathered} \Rightarrow \frac{1}{4}\left[\sin (36°+54°)–\sin (36°–54°)–\sin 54°\right] \\ \Rightarrow \frac{1}{4}\left[\sin 90°+\sin 18°–\sin 54°\right] \\ \Rightarrow \frac{1}{4}\left[1+\frac{\sqrt{5}–1}{4}–\frac{\sqrt{5}+1}{4}\right] \\ \Rightarrow \frac{1}{4}\left[\frac{4+\sqrt{5}-1-\sqrt{5}-1}{4}\right] \\ \Rightarrow \frac{1}{4}\left[\frac{4-2}{4}\right] \\ \Rightarrow \frac{1}{4}\left(\frac{2}{4}\right) \\ \Rightarrow \frac{1}{8} \end{gathered}$$

Hence, option C is correct.