Question

Cos(2arccos 1/2 - 3 arccos 0 - arccos (-1/2))

Answer 1

cos(2п/3-3П/2-2П/3)=сos(4П/6-9П/6-4П/6)=сos(-9П/6)=cos(-3П/2)=0

сначала считаем arccos 1/2, 0, (-1/2), потом приводим к общему знаменателю и считаем..

 

Answer 2

To evaluate the expression cos(2arccos(1/2) - 3arccos(0) - arccos(-1/2)), let's break it down step by step.

First, let's simplify the innermost arccos terms:

arccos(1/2) = π/3 (since cos(π/3) = 1/2)

arccos(0) = π/2 (since cos(π/2) = 0)

arccos(-1/2) = 2π/3 (since cos(2π/3) = -1/2)

Now, substitute these values back into the original expression:

cos(2arccos(1/2) - 3arccos(0) - arccos(-1/2)) = cos(2(π/3) - 3(π/2) - 2π/3)

Now, simplify the expression within the cosine function:

2(π/3) - 3(π/2) - 2π/3 = 2π/3 - 3π/2 - 2π/3 = -3π/2

Finally, evaluate the cosine of -3π/2:

cos(-3π/2) = cos(π/2) = 0

Therefore, cos(2arccos(1/2) - 3arccos(0) - arccos(-1/2)) = 0.