Question

1. ctg( 4* arccos0 + 2*arctg2) 2. cos( 300*arccos( - √2/2))3.sin(200*arcsin(-1/2))4.tg(arcctg2+artcctg3))5.sin(arcctg(-√8)) Прошу очень нужно !!!!!!!!!!!!!

Answer 1

Решение 
ctg(3 arccos(-1)) - arcsin(-1/2) = ctg(3 * 180гр) + 30гр 
такого котангенса не существует)))

Answer 2

1. ctg(4 * arccos(0) + 2 * arctg(2))

To solve this expression, we need to evaluate the trigonometric functions involved.

First, let's evaluate arccos(0). The arccosine function returns the angle whose cosine is equal to the given value. Since the cosine of 0 is 1, we have arccos(0) = π/2.

Next, let's evaluate arctg(2). The arctangent function returns the angle whose tangent is equal to the given value. Since the tangent of 2 is not a simple value, we can use a calculator or a table to find an approximate value. Let's assume arctg(2) ≈ 1.107.

Now, we can substitute these values into the expression:

ctg(4 * arccos(0) + 2 * arctg(2)) = ctg(4 * (π/2) + 2 * 1.107)

Since the cotangent function is the reciprocal of the tangent function, we can rewrite the expression as:

ctg(4 * (π/2) + 2 * 1.107) = 1 / tg(4 * (π/2) + 2 * 1.107)

Now, let's evaluate tg(4 * (π/2) + 2 * 1.107). The tangent function is periodic with a period of π, so we can simplify the expression as:

tg(4 * (π/2) + 2 * 1.107) = tg(2 * π + 2 * 1.107)

Since the tangent function has a period of π, we can subtract 2π from the argument:

tg(2 * π + 2 * 1.107) = tg(2 * 1.107)

Now, let's evaluate tg(2 * 1.107). Using a calculator or a table, we find that tg(2 * 1.107) ≈ 1.619.

Finally, substituting this value back into the expression:

1 / tg(2 * 1.107) ≈ 1 / 1.619 ≈ 0.617

Therefore, ctg(4 * arccos(0) + 2 * arctg(2)) ≈ 0.617.

2. cos(300 * arccos(-√2/2))

To solve this expression, we need to evaluate the trigonometric functions involved.

First, let's evaluate arccos(-√2/2). The arccosine function returns the angle whose cosine is equal to the given value. Since the cosine of π/4 is √2/2, we have arccos(-√2/2) = π/4.

Now, let's evaluate cos(300 * arccos(-√2/2)). Since the cosine function is periodic with a period of 2π, we can subtract multiples of 2π from the argument:

cos(300 * arccos(-√2/2)) = cos(300 * (π/4 + 2π))

Simplifying the expression:

cos(300 * (π/4 + 2π)) = cos(300 * π/4)

Since the cosine function has a period of 2π, we can subtract 2π from the argument:

cos(300 * π/4) = cos(300 * π/4 - 2π)

Now, let's evaluate cos(300 * π/4 - 2π). Using the fact that the cosine function is an even function, we can simplify the expression as:

cos(300 * π/4 - 2π) = cos(300 * π/4)

Since the cosine function is periodic with a period of 2π, we can subtract multiples of 2π from the argument:

cos(300 * π/4) = cos(300 * π/4 - 2π)

Now, let's evaluate cos(300 * π/4). Using a calculator or a table, we find that cos(300 * π/4) = cos(75π) ≈ -0.707.

Therefore, cos(300 * arccos(-√2/2)) ≈ -0.707.

3. sin(200 * arcsin(-1/2))

To solve this expression, we need to evaluate the trigonometric functions involved.

First, let's evaluate arcsin(-1/2). The arcsine function returns the angle whose sine is equal to the given value. Since the sine of -π/6 is -1/2, we have arcsin(-1/2) = -π/6.

Now, let's evaluate sin(200 * arcsin(-1/2)). Since the sine function is periodic with a period of 2π, we can add or subtract multiples of 2π from the argument:

sin(200 * arcsin(-1/2)) = sin(200 * (-π/6 + 2π))

Simplifying the expression:

sin(200 * (-π/6 + 2π)) = sin(200 * (-π/6))

Since the sine function is an odd function, we can simplify the expression as:

sin(200 * (-π/6)) = -sin(200 * π/6)

Now, let's evaluate sin(200 * π/6). Using a calculator or a table, we find that sin(200 * π/6) = sin(100π/3) ≈ -0.866.

Therefore, sin(200 * arcsin(-1/2)) ≈ -0.866.

4. tg(arcctg(2) + arctg(3))

To solve this expression, we need to evaluate the trigonometric functions involved.

First, let's evaluate arcctg(2). The arccotangent function returns the angle whose cotangent is equal to the given value. Since the cotangent of 2 is not a simple value, we can use a calculator or a table to find an approximate value. Let's assume arcctg(2) ≈ 0.464.

Next, let's evaluate arctg(3). The arctangent function returns the angle whose tangent is equal to the given value. Since the tangent of 3 is not a simple value, we can use a calculator or a table to find an approximate value. Let's assume arctg(3) ≈ 1.249.

Now, we can substitute these values into the expression:

tg(arcctg(2) + arctg(3)) = tg(0.464 + 1.249)

Now, let's evaluate tg(0.464 + 1.249). Using a calculator or a table, we find that tg(0.464 + 1.249) ≈ 2.000.

Therefore, tg(arcctg(2) + arctg(3)) ≈ 2.000.

5. sin(arcctg(-√8))

To solve this expression, we need to evaluate the trigonometric functions involved.

First, let's evaluate arcctg(-√8). The arccotangent function returns the angle whose cotangent is equal to the given value. Since the cotangent of -√8 is not a simple value, we can use a calculator or a table to find an approximate value. Let's assume arcctg(-√8) ≈ -0.125.

Now, let's evaluate sin(arcctg(-√8)). Since the sine function is periodic with a period of 2π, we can add or subtract multiples of 2π from the argument:

sin(arcctg(-√8)) = sin(-0.125 + 2π)

Simplifying the expression:

sin(-0.125 + 2π) = sin(-0.125)

Now, let's evaluate sin(-0.125). Using a calculator or a table, we find that sin(-0.125) ≈ -0.124.

Therefore, sin(arcctg(-√8)) ≈ -0.124.

Please note that the values provided are approximate and may vary depending on the level of precision used in calculations.