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Prove that,

${\sin ^3}30^\circ - {\sin ^3}18^\circ = {\sin ^2}18^\circ $

We have, ${\sin ^3}18^\circ  + {\sin ^2}18^\circ  = {\sin ^2}18^\circ (\sin 18^\circ  + 1)$

$ = {\left( {\frac{{\sqrt 5  - 1}}{4}} \right)^2}\left( {\frac{{\sqrt 5  - 1}}{4} + 1} \right)$

$ = \frac{{6 - 2\sqrt 5 }}{{16}} \times \frac{{\sqrt 5  + 3}}{4}$

$ = \frac{{3 - \sqrt 5 }}{8} \times \frac{{3 + \sqrt 5 }}{4}$

$ = \frac{{9 - 5}}{{8 \times 4}} = \frac{1}{8} = {\sin ^3}30^\circ $

$ \Rightarrow {\sin ^2}18^\circ  = {\sin ^3}30^\circ  - {\sin ^3}18^\circ $

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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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