Q62 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
$\cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ}=$
- A$-\frac{1}{8}$
- B$-\frac{1}{4}$
- Correct answerC$\frac{1}{8}$
- D$\frac{1}{4}$
Solution
$\begin{aligned} & \cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ} \\ = & \cos 6^{\circ} \sin 24^{\circ} \cos (90-18)^{\circ} \\ = & \frac{1}{2}\left[2 \cos 6^{\circ} \sin 24^{\circ}\right] \sin 18^{\circ} \\ = & \frac{1}{2}\left[\sin (6+24)^{\circ}-\sin (6-24)^{\circ}\right] \sin 18^{\circ}\end{aligned}$
$\begin{aligned} & =\frac{1}{2}\left[\sin 30^{\circ} \sin 18^{\circ}+\sin 18^{\circ} \sin 18^{\circ}\right] \\ & =\frac{1}{2}\left[\frac{1}{2}\left(\frac{\sqrt{5}-1}{4}\right)+\left(\frac{\sqrt{5}-1}{4}\right)^2\right]=\frac{1}{8}\end{aligned}$
Hence, the correct answer is option: 'C'.