5 arctg ( - √3) + arccos(-√2/2)

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вопрос

We first need to find the numerical values of arctg(-√3) and arccos(-√2/2).

Given that arctg(-√3) = arctan(-√3), we know that the arctan function outputs an angle whose tangent is equal to the input. In this case, we have:

tan(arctan(-√3)) = -√3

This means that the angle arctan(-√3) must be in the fourth quadrant where tangent is negative. Using the tangent half-angle identity tan(θ/2) = ±√((1 - cos(θ)) / (1 + cos(θ))), we can find the cosine of the angle:

cos(arctan(-√3)) = 1 / √(1 + (-√3)^2) = 1 / 2

Therefore, arctan(-√3) = arccos(1 / 2) = π / 3.

Now, let's evaluate arccos(-√2/2):

cos(arccos(-√2/2)) = -√2 / 2

This means that the angle arccos(-√2/2) must be in the second or third quadrant where cosine is negative. We can determine that this angle is -π/4.

Putting it all together, the expression simplifies to:

5 * (π/3) + (-π/4) = 5π/3 - π/4 = (20π - 3π) / 12 = 17π / 12

Therefore, the answer is 17π / 12.