Question Number 148513 by mathdanisur last updated on 28/Jul/21
$$\mathrm{6}\:+\:{log}_{\mathrm{2}} \:{sin}\mathrm{15}°\:-\:{log}_{\frac{\mathrm{1}}{\mathrm{2}}} {sin}\mathrm{75}°\:=\:? \\ $$
Answered by liberty last updated on 29/Jul/21
$$\chi\:=\:\mathrm{6}\:+\mathrm{log}\:_{\mathrm{2}} \mathrm{sin}\:\mathrm{15}°+\mathrm{log}\:_{\mathrm{2}} \mathrm{sin}\:\mathrm{75}° \\ $$$$\chi=\mathrm{6}+\mathrm{log}\:_{\mathrm{2}} \mathrm{sin}\:\mathrm{15}°+\mathrm{log}\:_{\mathrm{2}} \mathrm{cos}\:\mathrm{15}° \\ $$$$\chi=\mathrm{6}+\mathrm{log}\:_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{30}°\right) \\ $$$$\chi=\mathrm{6}+\mathrm{log}\:_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{6}+\mathrm{2}\left(−\mathrm{1}\right)=\:\mathrm{4} \\ $$
Commented by mathdanisur last updated on 29/Jul/21
$${Thank}\:{you}\:{Ser} \\ $$