arccos(-√3/2) Using the unit circle, we know that the cosine value is negative in the second and third quadrants. So, arccos(-√3/2) is in the second quadrant where the reference angle is π/6. Therefore, arccos(-√3/2) = π - π/6 = 5π/6.
arcsin(-√3/2) Using the unit circle, we know that the sine value is negative in the third and fourth quadrants. So, arcsin(-√3/2) is in the fourth quadrant where the reference angle is π/3. Therefore, arcsin(-√3/2) = -π/3.
Let's solve this step by step:
arccos(-√3/2)
Using the unit circle, we know that the cosine value is negative in the second and third quadrants. So, arccos(-√3/2) is in the second quadrant where the reference angle is π/6. Therefore, arccos(-√3/2) = π - π/6 = 5π/6.
arcsin(-√3/2)
Using the unit circle, we know that the sine value is negative in the third and fourth quadrants. So, arcsin(-√3/2) is in the fourth quadrant where the reference angle is π/3. Therefore, arcsin(-√3/2) = -π/3.
Adding both values:
5π/6 + (-π/3) = 5π/6 - π/3 = 5π/6 - 2π/6 = 3π/6 = π/2
Therefore, arccos(-√3/2) + arcsin(-√3/2) = π/2.