@bemath@ given { ((f(x)=log _2 (sin x)+log _2 (cos x))),((g(x)=log _2 (cos 2x)+log _2 (cos 4x))) :} find f((π/(48)))+g((π/(48))) .


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Question Number 106888 by bemath last updated on 07/Aug/20
$@bemath@ given { ((f(x)=log _2 (sin x)+log _2 (cos x))),((g(x)=log _2 (cos 2x)+log _2 (cos 4x))) :} find f((π/(48)))+g((π/(48))) .$
$@ bemath @$ $g iven {\begin{cases} f (x) = \log_{2} (\sin x) + \log_{2} (\cos x) \\ g (x) = \log_{2} (\cos 2 x) + \log_{2} (\cos 4 x) \end{cases}$ $find f (\frac{π}{48}) + g (\frac{π}{48}) .$
	Answered by Dwaipayan Shikari last updated on 07/Aug/20

	$f (x) = {l o g}_{2} (s i n 2 x) - 1$ $g (x) = {l o g}_{2} (c o s 2 x) + {l o g}_{2} (c o s 4 x)$ $f (x) + g (x) = {l o g}_{2} (s i n 2 x) + {l o g}_{2} (c o s 4 x) + l o g (c o s 2 x) - 1$ $f (x) + g (x) = {l o g}_{2} (s i n 4 x) + {l o g}_{2} (c o s 4 x) - 3$ $f (\frac{π}{48}) + g (\frac{π}{48}) = {l o g}_{2} (s i n 8 x) - 2 = l$ $= {l o g}_{2} (\frac{1}{2}) - 3 = - 4$
	Commented by Dwaipayan Shikari last updated on 07/Aug/20

	$O h i p u t c o s 4 x i n s t e a d o f s i n 4 x . t y p o$
	Commented by bemath last updated on 07/Aug/20

	$wrong sir . please check$
	Commented by Dwaipayan Shikari last updated on 07/Aug/20

	$T h a n k i n g f o r y o u r c o r r e c t i o n$
	Answered by bemath last updated on 07/Aug/20

	$@ bemath @$ $f (x) + g (x) = \log_{2} (\sin 2 x) + \log_{2} (\cos 2 x) + \log_{2} (\cos 4 x) - 1$ $= \log_{2} (\sin 4 x) + \log_{2} (\cos 4 x) - 2$ $= \log_{2} (\sin 8 x) - 3$ $f (\frac{π}{48}) + g (\frac{π}{48}) = \log_{2} (\sin \frac{π}{6}) - 3$ $= \log_{2} (\frac{1}{2}) - 3 = - 4$
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