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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=$
  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{4}$
  • ✓
    $\frac{1}{8}$
  • D
    $\frac{1}{16}$
✓

Answer

Correct option: C.
$\frac{1}{8}$
(C)
$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)$
$=\left(1+\cos \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)$$\ldots[\because \cos (\pi-\theta)=-\cos \theta]$
$=\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right)$
$=\sin ^2 \frac{\pi}{8} \sin ^2 \frac{3 \pi}{8}$
$=\frac{1}{4}\left(2 \sin \frac{\pi}{8} \cdot \sin \frac{3 \pi}{8}\right)^2$
$=\frac{1}{4}\left(\cos \frac{\pi}{4}-\cos \frac{\pi}{2}\right)^2$
$=\frac{1}{8}$

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