bemath-given-f-x-log-2-sin-x-log-2-cos-x-g-x-log-2-cos-2x-log-2-cos-4x-find-f-pi-48-g-pi-48-

Question Number 106888 by bemath last updated on 07/Aug/20

@ 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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