Question

Arccos(-√3/2)+arcsin√3/2​

Answer 1

Ответ: решение во вложении

Answer 2

To evaluate the expression arccos(-√3/2) + arcsin(√3/2), we can use the trigonometric identities for the inverse cosine (arccos) and inverse sine (arcsin) functions.

Recall that the arccos function gives the angle whose cosine is a given value, and the arcsin function gives the angle whose sine is a given value.

First, let's evaluate arccos(-√3/2):

arccos(-√3/2) = π - arccos(√3/2)

The reason for this is that the cosine function is negative in the second and third quadrants of the unit circle, and arccos always gives a value between 0 and π. Therefore, arccos(-√3/2) is equivalent to subtracting the value of arccos(√3/2) from π.

Now, let's evaluate arcsin(√3/2):

arcsin(√3/2) = π/2

The sine function is positive in the first and second quadrants, and arcsin always gives a value between -π/2 and π/2. Since √3/2 corresponds to the value of sin(π/3), the arcsin of √3/2 is π/3 or 60 degrees.

Now, substituting the values we found back into the expression:

arccos(-√3/2) + arcsin(√3/2) = (π - arccos(√3/2)) + (π/2)

Since arccos(√3/2) = π/3, we can simplify further:

(π - arccos(√3/2)) + (π/2) = (π - π/3) + (π/2) = (2π/3) + (π/2)

Adding the fractions:

(2π/3) + (π/2) = (4π/6) + (3π/6) = (7π/6)

Therefore, arccos(-√3/2) + arcsin(√3/2) equals (7π/6) or approximately 3.66519142918809.