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In acute angled $\Delta ABC$ prove that,

$sin A + sin B + sin C > cos A + cos B + cos C$

In acute angled triangle, $\cos \frac{A}{2} > \sin \frac{A}{2}$

So, $\sin \frac{A}{2}.\cos \frac{A}{2} > \sin \frac{A}{2}.\sin \frac{A}{2}$

Hence, $2\sin \frac{A}{2}\cos \frac{A}{2} > 2{\sin ^2}\frac{A}{2}$

Thus, $\sin A > 1 - \cos A$

Likewise, $\sin B > 1 - \cos B$ & $\sin C > 1 - \cos C$

On addition, $\sin A + \sin B + \sin C > 3 - (\cos A + \cos B + \cos C)$ ........(!)

In any triangle, $\cos A + \cos B + \cos C \le \frac{3}{2}$*

$\therefore 2(\cos A + \cos B + \cos C) \le 3$

$ \Rightarrow \cos A + \cos B + \cos C \le 3 - (\cos A + \cos B + \cos C)$ ........(!!)

From (!) & (!!), $\sin A + \sin B + \sin C > \cos A + \cos B + \cos C$

*Please refer textbook or study material for the proof.

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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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