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Prove That,

$\cos \frac{\pi }{7}-\cos \frac{{2\pi }}{7}+\cos \frac{{3\pi}}{7}=\frac{1}{2}$

L.H.S.=$\cos \frac{\pi }{7} - \cos \frac{{2\pi }}{7} + \cos \frac{{3\pi }}{7}$

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

= $\cos \frac{\pi }{7} + \cos \frac{{3\pi }}{7} + \cos \frac{{5\pi }}{7}$

= $\cos \left( {\frac{\pi }{7} + \frac{{3 - 1}}{2}.\frac{{2\pi }}{7}} \right).\frac{{\sin \left( {3.\frac{{2\pi /7}}{2}} \right)}}{{\sin \left( {\frac{{2\pi /7}}{2}} \right)}}$

= $\cos \left( {\frac{\pi }{7} + \frac{{2\pi }}{7}} \right).\frac{{\sin \frac{{3\pi }}{7}}}{{\sin \frac{\pi }{7}}}$

= $\frac{{\cos \frac{{3\pi }}{7}\sin \frac{{3\pi }}{7}}}{{\sin \frac{\pi }{7}}} = \frac{{2\cos \frac{{3\pi }}{7}\sin \frac{{3\pi }}{7}}}{{2\sin \frac{\pi }{7}}}$

= $\frac{{\sin \frac{{6\pi }}{7}}}{{2\sin \frac{\pi }{7}}} = \frac{{\sin \left( {\pi - \frac{\pi }{7}} \right)}}{{2\sin \frac{\pi }{7}}} = \frac{1}{2}$

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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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