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Question Number 148513 by mathdanisur last updated on 28/Jul/21

6 + log_2  sin15° - log_(1/2) sin75° = ?

$$\mathrm{6}\:+\:{log}_{\mathrm{2}} \:{sin}\mathrm{15}°\:-\:{log}_{\frac{\mathrm{1}}{\mathrm{2}}} {sin}\mathrm{75}°\:=\:? \\ $$

Answered by liberty last updated on 29/Jul/21

χ = 6 +log _2 sin 15°+log _2 sin 75°  χ=6+log _2 sin 15°+log _2 cos 15°  χ=6+log _2 ((1/2)sin 30°)  χ=6+log _2 ((1/2))^2 =6+2(−1)= 4

$$\chi\:=\:\mathrm{6}\:+\mathrm{log}\:_{\mathrm{2}} \mathrm{sin}\:\mathrm{15}°+\mathrm{log}\:_{\mathrm{2}} \mathrm{sin}\:\mathrm{75}° \\ $$$$\chi=\mathrm{6}+\mathrm{log}\:_{\mathrm{2}} \mathrm{sin}\:\mathrm{15}°+\mathrm{log}\:_{\mathrm{2}} \mathrm{cos}\:\mathrm{15}° \\ $$$$\chi=\mathrm{6}+\mathrm{log}\:_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{30}°\right) \\ $$$$\chi=\mathrm{6}+\mathrm{log}\:_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{6}+\mathrm{2}\left(−\mathrm{1}\right)=\:\mathrm{4} \\ $$

Commented by mathdanisur last updated on 29/Jul/21

Thank you Ser

$${Thank}\:{you}\:{Ser} \\ $$

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