The largest interval in which the function f(x)=3sin [ square root (pi^2/16 -x^2) ]

Solution:     The given function is  defined if

$(\frac{\pi^{2}}{16}-x^{2})\geq 0\Rightarrow -\frac{\pi}{4}\leq x\leq \frac{\pi}{4}.$

$\because$     $f(-\frac{\pi}{4})=0;f(\frac{\pi}{4})=0;3\sin (\frac{\pi}{4})=\frac{3}{\sqrt{2}}$

$\therefore$         $f$  takes values in the interval    $[0,\frac{3}{\sqrt{2}}].$

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