Question

4 arccos(-√2/2)+arctg(-1)+arcsin(-1)=

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

To find the value of the expression 4 arccos(-√2/2) + arctg(-1) + arcsin(-1), we need to evaluate each individual term first:

1. arccos(-√2/2): The value of arccos(-√2/2) is the angle whose cosine is -√2/2. Since the cosine function is positive in the first and fourth quadrants, and -√2/2 is the cosine value of π/4 (45 degrees) in the second quadrant, the principal value of arccos(-√2/2) is π - π/4 = 3π/4 (135 degrees).

2. arctg(-1): The value of arctg(-1) is the angle whose tangent is -1. This corresponds to -π/4 (-45 degrees).

3. arcsin(-1): The value of arcsin(-1) is the angle whose sine is -1. Since the sine function is negative in the third and fourth quadrants, and -1 is the sine value of -π/2 (-90 degrees) in the fourth quadrant, the principal value of arcsin(-1) is -π/2 (-90 degrees).

Now, let's calculate the sum:

4 arccos(-√2/2) + arctg(-1) + arcsin(-1) = 4 * (3π/4) + (-π/4) + (-π/2)

Next, simplify the expression:

4 * (3π/4) = 3π (-π/4) + (-π/2) = -3π/4

Now, put everything together:

3π + (-3π/4) = (12π - 3π) / 4 = 9π / 4

So, the final value of the expression is 9π / 4, which is approximately 7.06858.