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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

	$T h a n k y o u S e r$
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