Question

3arctg(-1/v3)+2arccos(-v3/2)

Answer 1

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Answer 2

To simplify the expression 3arctg(-1/v3) + 2arccos(-v3/2), we need to find the values of the individual trigonometric functions and then perform the arithmetic.

Let's start with each part of the expression:

1. arctg(-1/v3): This represents the arctangent of (-1/v3).

The arctangent function is the inverse of the tangent function and gives us the angle whose tangent is a given value. In this case, we want to find the angle whose tangent is -1/v3.

arctg(-1/v3) = -π/6

2. arccos(-v3/2): This represents the arccosine of (-v3/2).

The arccosine function is the inverse of the cosine function and gives us the angle whose cosine is a given value. In this case, we want to find the angle whose cosine is -v3/2.

arccos(-v3/2) = 5π/6

Now, we can substitute these values back into the original expression:

3arctg(-1/v3) + 2arccos(-v3/2) = 3(-π/6) + 2(5π/6) = -π/2 + 5π/3

To simplify further, we find a common denominator for -π/2 and 5π/3, which is 6:

-π/2 = -3π/6

5π/3 = 10π/6

So the expression becomes:

-3π/6 + 10π/6 = 7π/6

Therefore, the simplified value of the expression 3arctg(-1/v3) + 2arccos(-v3/2) is 7π/6.