Question

The numerical value of $sin{12}^{\circ }sin{48}^{\circ }\cdot sin{54}^{\circ }$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{16}$
• D. $\frac{1}{8}$

Answer 1

The correct option is D $\frac{1}{8}$

$\sin {12}^{\circ }\sin {48}^{\circ }=\frac{1}{2}\left(2\sin {12}^{\circ }\sin {48}^{\circ }\right)$

$=\frac{1}{2}\left(\cos \right({48}^{\circ }-{12}^{\circ })-\cos ({48}^{\circ }+{12}^{\circ }\left)\right)$

$=\frac{1}{2}(\cos {36}^{\circ }-\cos {60}^{\circ })$

$=\frac{1}{2}(\frac{\sqrt{5}+1}{4}-\frac{1}{2})$

$=\frac{\sqrt{5}-1}{8}$

$\sin {54}^{\circ }=\cos ({90}^{\circ }-{54}^{\circ })=\cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}$

$\sin {12}^{\circ }\sin {48}^{\circ }\sin {54}^{\circ }=\frac{\sqrt{5}-1}{8}\times \frac{\sqrt{5}+1}{4}=\frac{(\sqrt{5}{)}^2-1^2}{32}=\frac{5-1}{32}=\frac{1}{8}$