Question

7cos(arctg √3/3+arcsin(-√3/2) ).

Answer 1

7cos(пи/6-пи/3))=7cos(-пи/6)=7cos(пи/6)=7*r3/2=3.5r3.  r- обозначение кв.корня

Answer 2

To simplify the expression 7cos(arctg(√3/3) + arcsin(-√3/2)), we can use the trigonometric identities and properties to break it down step by step.

1. First, let's find the values of the trigonometric functions for the angles involved:

   • arctg(√3/3) is the same as arctan(√3/3). To find the reference triangle for this angle, we consider a right triangle with an angle θ and sides √3 and 3. Using the Pythagorean theorem, we find the length of the hypotenuse to be 2. Therefore, sin(θ) = √3/2, and cos(θ) = 1/2.
   • arcsin(-√3/2) is an angle whose sine is -√3/2. This corresponds to an angle of -π/3 or -60 degrees.

2. Now, let's rewrite the expression with these values: 7cos(arctg(√3/3) + arcsin(-√3/2)) = 7cos(θ + (-π/3))

3. Use the sum of angles formula for cosine: cos(θ + (-π/3)) = cos(θ)cos(-π/3) - sin(θ)sin(-π/3)

4. Plug in the values we found earlier: = (1/2)cos(-π/3) - (√3/2)sin(-π/3)

5. We know that cos(-π/3) = cos(π/3) and sin(-π/3) = -sin(π/3). So, we have: = (1/2)cos(π/3) + (√3/2)sin(π/3)

6. Using the values of sin(π/3) and cos(π/3), which can be found from a 30-60-90 triangle, where the hypotenuse is 2: sin(π/3) = √3/2 cos(π/3) = 1/2

7. Substitute these values into the expression: = (1/2)(1/2) + (√3/2)(√3/2)

8. Simplify further: = 1/4 + 3/4

9. Add the fractions: = 4/4

10. Finally, simplify to the final result: = 1

So, 7cos(arctg(√3/3) + arcsin(-√3/2)) simplifies to 1.