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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
$\sin 12^{\circ} \sin 24^{\circ} \sin 48^{\circ} \sin 84^{\circ}=$
  • ✓
    $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$
  • B
    $\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}$
  • C
    $\frac{3}{15}$
  • D
    $\frac{5}{16}$
✓

Answer

Correct option: A.
$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$
(A)
$\sin 12^{\circ} \sin 24^{\circ} \sin 48^{\circ} \sin 84^{\circ}$
$=\frac{1}{4}\left(2 \sin 12^{\circ} \sin 48^{\circ}\right)\left(2 \sin 24^{\circ} \sin 84^{\circ}\right)$
$=\frac{1}{2}\left(\cos 36^{\circ}-\cos 60^{\circ}\right)\left(\cos 60^{\circ}-\cos 108^{\circ}\right)$
$=\frac{1}{4}\left(\cos 36^{\circ}-\frac{1}{2}\right)\left(\frac{1}{2}+\sin 18^{\circ}\right)$
$=\frac{1}{4}\left\{\frac{1}{4}(\sqrt{5}+1)-\frac{1}{2}\right\}\left\{\frac{1}{2}+\frac{1}{4}(\sqrt{5}-1)\right\}=\frac{1}{16}$
Consider, $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}$
$=\frac{1}{2}\left[\cos \left(60^{\circ}-20^{\circ}\right) \cos 20^{\circ} \cos \left(60^{\circ}+20^{\circ}\right)\right]$
$\cos \theta \cos \left(60^{\circ}-\theta\right) \cos \left(60^{\circ}+\theta\right)=\frac{1}{4} \cos 3 \theta$
$=\frac{1}{2}\left[\frac{1}{4} \cos 3\left(20^{\circ}\right)\right]$
$=\frac{1}{8} \cos 60^{\circ}=\frac{1}{8} \times \frac{1}{2}=\frac{1}{16}$
$\therefore$ option (A) is the correct answer.

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