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${\sin ^2}{1^\circ } + {\sin ^2}{2^\circ } + ............ + {\sin ^2}{180^\circ } = ?$

The given expression can be written as,

$\frac{1}{2}\left[ {(1 - \cos {2^\circ }) + (1 - \cos {4^\circ }) + ................ + (1 - \cos {{360}^\circ })} \right]$

$ = \frac{1}{2}\left[ {(1 + 1 + ........ + 1) - (\cos {2^\circ } + \cos {4^\circ } + ........... + \cos {{360}^\circ })} \right]$

There are 180 terms in the series.

$ = \frac{1}{2}\left[ {180 - \frac{{\cos \left( {{2^\circ } + \frac{{180 - 1}}{2} \times {2^\circ }} \right)\sin \left( {\frac{{180 \times {2^\circ }}}{2}} \right)}}{{\sin \left( {\frac{{{2^\circ }}}{2}} \right)}}} \right]$

$ = \frac{1}{2}\left( {180 - \frac{{\cos {{181}^\circ }\sin {{180}^\circ }}}{{\sin {1^\circ }}}} \right) = 90$

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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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