$\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}} \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) = $
- [IIT 1984]
- A
$\frac{1}{2}$
- B
$\frac{1}{4}$
- C
$\frac{1}{8}$
- D
$\frac{1}{{16}}$
Similar Questions
यदि $\cos \,(\theta - \alpha ) = a,\,\,\sin \,(\theta - \beta ) = b,\,\,$ हो, तब ${\cos ^2}(\alpha - \beta ) + 2ab\,\sin \,(\alpha - \beta )$ बराबर है
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$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
- [IIT 1974]
$\cos \alpha .\sin (\beta - \gamma ) + \cos \beta .\sin (\gamma - \alpha ) + \cos \gamma .\sin (\alpha - \beta ) = $
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यदि $\tan \alpha = \frac{1}{7}$ तथा $\sin \beta = \frac{1}{{\sqrt {10} }}\left( {0 < \alpha ,\,\beta < \frac{\pi }{2}} \right)$, तब $2\beta $ बराबर है
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