Question

Arcsin(-1)+arccos1+arctg корень3/3+arcctg(-корень3) Пожалуйста помогите!

Answer 1

Arcsin(-1)+arccos1+arctg✓3/3+arcctg(-3)=-p/2+0+p/6+5p/6=-p/2+5p/6=p/3

Answer 2

Calculation of Trigonometric Functions

To calculate the values of the given trigonometric functions, let's go through each one step by step:

1. Arcsin(-1): - The arcsin function returns the angle whose sine is equal to the given value. - The value -1 is the minimum value for the sine function, which occurs at -90 degrees or -π/2 radians. - Therefore, the arcsin(-1) is equal to -π/2.

2. Arccos(1): - The arccos function returns the angle whose cosine is equal to the given value. - The value 1 is the maximum value for the cosine function, which occurs at 0 degrees or 0 radians. - Therefore, the arccos(1) is equal to 0.

3. Arctg(√3/3): - The arctg function returns the angle whose tangent is equal to the given value. - To calculate arctg(√3/3), we need to find the angle whose tangent is equal to √3/3. - By taking the inverse tangent of √3/3, we find that the angle is approximately π/6 or 30 degrees.

4. Arcctg(-√3): - The arcctg function returns the angle whose cotangent is equal to the given value. - To calculate arcctg(-√3), we need to find the angle whose cotangent is equal to -√3. - By taking the inverse cotangent of -√3, we find that the angle is approximately -π/3 or -60 degrees.

Please let me know if there's anything else I can help you with!