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3. trignometrical ratios,functions and identities — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS3. trignometrical ratios,functions and identities1 Mark
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
(c) $\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right)$

$ = \left( {1 + \cos \frac{\pi }{8} + \cos \frac{{7\pi }}{8} + \cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)$

$\left( {1 + \cos \frac{{5\pi }}{8} + \cos \frac{{3\pi }}{8} + \cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$

$ = \left( {1 + \cos \frac{\pi }{8} - \cos \frac{\pi }{8} + \cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)$

$\left( {1 + \cos \frac{{5\pi }}{8} - \cos \frac{{5\pi }}{8} + \cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$

$ = \left( {1 + \cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$

$ = \frac{1}{4}\,\,\left( {2 + 2\cos \frac{\pi }{8}\cos \frac{{7\pi }}{8}} \right)\,\,\left( {2 + 2\cos \frac{{3\pi }}{8}\cos \frac{{5\pi }}{8}} \right)$
$ = \frac{1}{4}\left( {2 + \cos \frac{{3\pi }}{4} + \cos \pi } \right)\left( {2 + \cos \frac{\pi }{4} + \cos \pi } \right)$

$ = \frac{1}{4}\,\left( {1 + \cos \frac{{3\pi }}{4}} \right)\,\left( {1 + \cos \frac{\pi }{4}} \right) = \frac{1}{4}\left( {1 - \cos \frac{\pi }{4}} \right)\,\left( {1 + \cos \frac{\pi }{4}} \right)$

$ = \frac{1}{4}\left( {1 - {{\cos }^2}\frac{\pi }{4}} \right) = \frac{1}{4}\left( {1 - \frac{1}{2}} \right) = \frac{1}{8}$.

Aliter : $\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)$

$ = \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)$

$ = \left( {1 - {{\cos }^2}\frac{\pi }{8}} \right){\rm{ }}\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}.\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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