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:

Answer 3

To solve this expression, we can simplify each logarithm term using logarithmic identities and then combine them.

First, let's simplify each term:

1. log2(ctg60): The cotangent function (ctg) is the reciprocal of the tangent function. We know that tan(60°) = √3, so ctg(60°) = 1 / tan(60°) = 1 / √3.

   Now we can rewrite log2(ctg60) as log2(1 / √3). Using the logarithmic identity logb(a/b) = logb(a) - logb(b), we can rewrite it as log2(1) - log2(√3). The logarithm of 1 to any base is always 0, so log2(1) = 0. We are left with -log2(√3).

2. log2(sin60): The sine function of 60° is √3 / 2.

   Now we can rewrite log2(sin60) as log2(√3 / 2). Using the logarithmic identity logb(a/b) = logb(a) - logb(b), we can rewrite it as log2(√3) - log2(2). The logarithm of 2 to the base 2 is 1, so log2(2) = 1. We are left with log2(√3) - 1.

Finally, we can combine the two simplified terms:

-log2(√3) + log2(√3) - 1.

The two logarithmic terms cancel each other out, leaving us with:

-1.

Therefore, log2(ctg60) + log2(sin60) = -1.