6-log-2-sin15-log-1-2-sin75-

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

6 + {l o g}_{2} s i n 15 ° - {l o g}_{\frac{1}{2}} s i n 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} (\frac{1}{2} \sin 30 °)

χ = 6 + \log_{2} {(\frac{1}{2})}^{2} = 6 + 2 (- 1) = 4

Commented by mathdanisur last updated on 29/Jul/21

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