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Trigonometrical equations — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometrical equations1 Mark
MCQ
If $L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ and $M=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right),$ then 
  • ✓
    $M =\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$
  • B
    $L =\frac{1}{4 \sqrt{2}}-\frac{1}{4} \cos \frac{\pi}{8}$
  • C
    $M =\frac{1}{4 \sqrt{2}}+\frac{1}{4} \cos \frac{\pi}{8}$
  • D
    $L =-\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$
✓

Answer

Correct option: A.
$M =\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$
a
$L =\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$

$\left(\because \sin ^{2} \theta=\frac{1-\cos 2 \theta}{2}\right)$

$\Rightarrow L =\left(\frac{1-\cos (\pi / 8)}{2}\right)-\left(\frac{1-\cos (\pi / 4)}{2}\right)$

$L =\frac{1}{2}\left[\cos \left(\frac{\pi}{4}\right)-\cos \left(\frac{\pi}{8}\right)\right]$

$L =\frac{1}{2 \sqrt{2}}-\frac{1}{2} \cos \left(\frac{\pi}{8}\right)$

$M =\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$

$M =\frac{1+\cos (\pi / 8)}{2}-\frac{1-\cos (\pi / 4)}{2}$

$M =\frac{1}{2} \cos \left(\frac{\pi}{8}\right)+\frac{1}{2 \sqrt{2}}$

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