Question Number 106888 by bemath last updated on 07/Aug/20
$$@\mathrm{bemath}@ \\ $$$$\mathfrak{g}\mathrm{iven}\:\begin{cases}{\mathrm{f}\left(\mathrm{x}\right)=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\mathrm{x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{x}\right)}\\{\mathrm{g}\left(\mathrm{x}\right)=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{2x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{4x}\right)}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{f}\left(\frac{\pi}{\mathrm{48}}\right)+\mathrm{g}\left(\frac{\pi}{\mathrm{48}}\right)\:. \\ $$
Answered by Dwaipayan Shikari last updated on 07/Aug/20
$${f}\left({x}\right)={log}_{\mathrm{2}} \left({sin}\mathrm{2}{x}\right)−\mathrm{1} \\ $$$${g}\left({x}\right)={log}_{\mathrm{2}} \left({cos}\mathrm{2}{x}\right)+{log}_{\mathrm{2}} \left({cos}\mathrm{4}{x}\right) \\ $$$${f}\left({x}\right)+{g}\left({x}\right)={log}_{\mathrm{2}} \left({sin}\mathrm{2}{x}\right)+{log}_{\mathrm{2}} \left({cos}\mathrm{4}{x}\right)+{log}\left({cos}\mathrm{2}{x}\right)−\mathrm{1} \\ $$$${f}\left({x}\right)+{g}\left({x}\right)={log}_{\mathrm{2}} \left({sin}\mathrm{4}{x}\right)+{log}_{\mathrm{2}} \left({cos}\mathrm{4}{x}\right)−\mathrm{3} \\ $$$${f}\left(\frac{\pi}{\mathrm{48}}\right)+{g}\left(\frac{\pi}{\mathrm{48}}\right)={log}_{\mathrm{2}} \left({sin}\mathrm{8}{x}\right)−\mathrm{2}={l} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{3}=−\mathrm{4} \\ $$$$ \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 07/Aug/20
$${Oh}\:{i}\:{put}\:{cos}\mathrm{4}{x}\:{instead}\:{of}\:{sin}\mathrm{4}{x}.{typo} \\ $$
Commented by bemath last updated on 07/Aug/20
$$\mathrm{wrong}\:\mathrm{sir}.\:\mathrm{please}\:\mathrm{check} \\ $$
Commented by Dwaipayan Shikari last updated on 07/Aug/20
$${Thanking}\:\:{for}\:{your}\:{correction} \\ $$
Answered by bemath last updated on 07/Aug/20
$$\:\:\:\:\:@\mathrm{bemath}@ \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{g}\left(\mathrm{x}\right)=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\mathrm{2x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{2x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{4x}\right)−\mathrm{1} \\ $$$$=\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\mathrm{4x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{4x}\right)−\mathrm{2} \\ $$$$=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\mathrm{8x}\right)−\mathrm{3} \\ $$$$\mathrm{f}\left(\frac{\pi}{\mathrm{48}}\right)+\mathrm{g}\left(\frac{\pi}{\mathrm{48}}\right)=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right)−\mathrm{3} \\ $$$$=\mathrm{log}\:_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{3}=−\mathrm{4} \\ $$