Question

-2arcsin(-√3/2)+6arccos(-√2/2)-9arctg(√3/3)

Answer 1

$arcsin (- \sqrt{3} \div 2) = - \pi \div 3$
$arccos( - \sqrt{2} \div 2) = \pi \div 4$
$arctg( \sqrt{3} \div 3 )= п/6$
$\frac{2\pi}{3} + \frac{3\pi}{2} - 3п/2$
кароч ответ 2п/3
но проверь лучше

Answer 2

To simplify the expression, let's start by evaluating each trigonometric function separately:

1. arcsin(-√3/2): The arcsine function returns an angle whose sine is equal to the given value. The sine function is positive in the second and fourth quadrants. Since -√3/2 is negative and lies between -1 and 0, the arcsine function will give an angle in the fourth quadrant.

Using the fact that sin(π/3) = √3/2, we can write:

arcsin(-√3/2) = -arcsin(√3/2) = -π/3

2. arccos(-√2/2): The arccosine function returns an angle whose cosine is equal to the given value. The cosine function is positive in the first quadrant and negative in the second quadrant. Since -√2/2 is negative and lies between -1 and 0, the arccosine function will give an angle in the second quadrant.

Using the fact that cos(π/4) = √2/2, we can write:

arccos(-√2/2) = π - arccos(√2/2) = π - π/4 = 3π/4

3. arctg(√3/3): The arctangent function returns an angle whose tangent is equal to the given value. The tangent function is positive in the first and third quadrants. Since √3/3 is positive and lies between 0 and 1, the arctangent function will give an angle in the first quadrant.

Using the fact that tan(π/6) = √3/3, we can write:

arctg(√3/3) = π/6

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

-2arcsin(-√3/2) + 6arccos(-√2/2) - 9arctg(√3/3)

= -2*(-π/3) + 6*(3π/4) - 9*(π/6)

= 2π/3 + 9π/2 - 3π/2

Now, to combine the terms with π:

= (2π + 9π - 3π) / 3

= 8π / 3

So, the simplified value of the expression is 8π / 3.