Question

log2 ctg60 + log2 sin60 = ?

Answer 1

Ответ:

$log_{2}\cot60° + log_{2}\sin60° = \\ = log_{2}\frac{ \sqrt{3} }{3} + log_{2}\frac{ \sqrt{3} }{2} = \\ = log_{2}( \frac{ \sqrt{3} }{3} \times \frac{ \sqrt{3} }{2} ) = log_{2}\frac{3}{6} = \\ = log_{2} \frac{1}{ 2 } = log_{2} {2}^{ - 1} = \\ = - 1 \times log_{2}(2) = - 1$

Answer 2

Ответ:

log2 ctg60 + log2 sin60 = log2(ctg60*sin60)=log(cos60)=log2(1/2)=-1

Пошаговое объяснение: