Question

Arcsin(-1/2)+arccos(-√2/2)+arctg0

Answer 1

Решение задания приложено

Answer 2

I hope this helps you

Answer 3

To calculate the given expression, let's find the values of each individual term:

1. arcsin(-1/2): The arcsine function (arcsin) returns an angle whose sine is equal to the input. The arcsin function is defined in the range [-π/2, π/2].

Since sin(-π/6) = -1/2, the value of arcsin(-1/2) is -π/6.

2. arccos(-√2/2): The arccosine function (arccos) returns an angle whose cosine is equal to the input. The arccos function is defined in the range [0, π].

Since cos(π/4) = √2/2, the value of arccos(-√2/2) is π/4.

3. arctan(0): The arctangent function (arctan) returns an angle whose tangent is equal to the input. The arctan function is defined in the range [-π/2, π/2].

Since tan(0) = 0, the value of arctan(0) is 0.

Now, let's sum up the values:

arcsin(-1/2) + arccos(-√2/2) + arctan(0)

= -π/6 + π/4 + 0

To perform the addition, we need to find a common denominator:

= (-2π + 3π) / 12

= π / 12

So, the final value of the expression is π / 12.