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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))) .

$$@\mathrm{bemath}@ \\ $$$$\mathfrak{g}\mathrm{iven}\:\begin{cases}{\mathrm{f}\left(\mathrm{x}\right)=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\mathrm{x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{x}\right)}\\{\mathrm{g}\left(\mathrm{x}\right)=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{2x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{4x}\right)}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{f}\left(\frac{\pi}{\mathrm{48}}\right)+\mathrm{g}\left(\frac{\pi}{\mathrm{48}}\right)\:. \\ $$

Answered by Dwaipayan Shikari last updated on 07/Aug/20

f(x)=log_2 (sin2x)−1 g(x)=log_2 (cos2x)+log_2 (cos4x) f(x)+g(x)=log_2 (sin2x)+log_2 (cos4x)+log(cos2x)−1 f(x)+g(x)=log_2 (sin4x)+log_2 (cos4x)−3 f((π/(48)))+g((π/(48)))=log_2 (sin8x)−2=l =log_2 ((1/2))−3=−4

$${f}\left({x}\right)={log}_{\mathrm{2}} \left({sin}\mathrm{2}{x}\right)−\mathrm{1} \\ $$$${g}\left({x}\right)={log}_{\mathrm{2}} \left({cos}\mathrm{2}{x}\right)+{log}_{\mathrm{2}} \left({cos}\mathrm{4}{x}\right) \\ $$$${f}\left({x}\right)+{g}\left({x}\right)={log}_{\mathrm{2}} \left({sin}\mathrm{2}{x}\right)+{log}_{\mathrm{2}} \left({cos}\mathrm{4}{x}\right)+{log}\left({cos}\mathrm{2}{x}\right)−\mathrm{1} \\ $$$${f}\left({x}\right)+{g}\left({x}\right)={log}_{\mathrm{2}} \left({sin}\mathrm{4}{x}\right)+{log}_{\mathrm{2}} \left({cos}\mathrm{4}{x}\right)−\mathrm{3} \\ $$$${f}\left(\frac{\pi}{\mathrm{48}}\right)+{g}\left(\frac{\pi}{\mathrm{48}}\right)={log}_{\mathrm{2}} \left({sin}\mathrm{8}{x}\right)−\mathrm{2}={l} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{3}=−\mathrm{4} \\ $$$$ \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 07/Aug/20

Oh i put cos4x instead of sin4x.typo

$${Oh}\:{i}\:{put}\:{cos}\mathrm{4}{x}\:{instead}\:{of}\:{sin}\mathrm{4}{x}.{typo} \\ $$

Commented by bemath last updated on 07/Aug/20

wrong sir. please check

$$\mathrm{wrong}\:\mathrm{sir}.\:\mathrm{please}\:\mathrm{check} \\ $$

Commented by Dwaipayan Shikari last updated on 07/Aug/20

Thanking for your correction

$${Thanking}\:\:{for}\:{your}\:{correction} \\ $$

Answered by bemath last updated on 07/Aug/20

@bemath@ f(x)+g(x)=log _2 (sin 2x)+log _2 (cos 2x)+log _2 (cos 4x)−1 = log _2 (sin 4x)+log _2 (cos 4x)−2 =log _2 (sin 8x)−3 f((π/(48)))+g((π/(48)))=log _2 (sin (π/6))−3 =log _2 ((1/2))−3=−4

$$\:\:\:\:\:@\mathrm{bemath}@ \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{g}\left(\mathrm{x}\right)=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\mathrm{2x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{2x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{4x}\right)−\mathrm{1} \\ $$$$=\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\mathrm{4x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{4x}\right)−\mathrm{2} \\ $$$$=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\mathrm{8x}\right)−\mathrm{3} \\ $$$$\mathrm{f}\left(\frac{\pi}{\mathrm{48}}\right)+\mathrm{g}\left(\frac{\pi}{\mathrm{48}}\right)=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right)−\mathrm{3} \\ $$$$=\mathrm{log}\:_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{3}=−\mathrm{4} \\ $$

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