3 arcsin(- корень 3/2)+2 arccos (-1/2)+arctg корень 3/3

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вопрос

There seems to be a mistake in the given expression since the range of arcsin is [-π/2, π/2] and the range of arccos is [0, π]. Here is the corrected version:

3 arcsin(-√3/2) + 2 arccos(-1/2) + arctan(√3/3)

Using trigonometric identities, we can evaluate each term:

arcsin(-√3/2):
Since sin(-π/3) = -√3/2, we have arcsin(-√3/2) = -π/3

arccos(-1/2):
Since cos(2π/3) = -1/2, we have arccos(-1/2) = 2π/3

arctan(√3/3):
Since tan(π/6) = √3/3, we have arctan(√3/3) = π/6

Substitute the values back into the expression:

3 (-π/3) + 2 (2π/3) + (π/6)
= -π + 4π/3 + π/6
= -6π/6 + 8π/6 + π/6
= 3π/6
= π/2

Therefore, the simplified value of the expression is π/2.