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photoshop1234 [79]
3 years ago
7

Which of the following points below represents (- 2, - (5pi)/4) the polar grid?

Mathematics
1 answer:
Maru [420]3 years ago
7 0

Answer:

Well, when we have a point in rectangular coordinates (x, y) and we want to write it in polar coordinates (r, θ) the rule we use is:

r = √(x^2 + y^2)

θ = Atg(Iy/xI) + 90°*(n - 1)

where n is number of the quadrant where our point is.

For example, if we are on the second quadrant, we use n = 2.

Now, if instead, we have the point (r, θ) and we want to rewrite it in rectangular coordinates, then the transformation is:

x = r*cos(θ)

y = r*sin(θ)

Now we have the point (- 2, -(5pi)/4) because the first part is negative, this number can not be in polar coordinates (r can not be negative) then we have:

x = -2

y = -(5*pi)/4

Using the first relation, we can find that:

r = √( (-2)^2 + (-5*3.14/4)^2) = 4.4

(rememer that the point (- 2, -(5pi)/4) is on the third quadrant, then we will use n = 3.)

θ = Atg(-(5*pi)/4/-2) + 90°*(3 - 1)= Atg((5*pi)/2) + 180° = 82.7° + 180° = 262.7°

Then the point that represents  (- 2, - (5pi)/4) in polar coordinates is the point:

(4.4, 262.7°)

If instead of degrees, the angle part is written in radians, we have:

180° = 3.14 radians.

262.7° = x radians.

then:

x/3.14 = (262.7)/180

x =  3.14*(262.7)/180 = 4.58 radians.

then the point will be:

(4.4, 4.58 radians)

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1) cos (θ / 2) = √[(1 + cos θ) / 2], sin (θ / 2) = √[(1 - cos θ) / 2], tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) (x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°). The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) The <em>linear</em> function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

<h3>How to apply trigonometry on deriving formulas and transforming points</h3>

1) The following <em>trigonometric</em> formulae are used to derive the <em>half-angle</em> formulas:

sin² θ / 2 + cos² θ / 2 = 1                      (1)

cos θ = cos² (θ / 2) - sin² (θ / 2)           (2)

First, we derive the formula for the sine of a <em>half</em> angle:

cos θ = 2 · cos² (θ / 2) - 1

cos² (θ / 2) = (1 + cos θ) / 2

cos (θ / 2) = √[(1 + cos θ) / 2]

Second, we derive the formula for the cosine of a <em>half</em> angle:

cos θ = 1 - 2 · sin² (θ / 2)

2 · sin² (θ / 2) = 1 - cos θ

sin² (θ / 2) = (1 - cos θ) / 2

sin (θ / 2) = √[(1 - cos θ) / 2]

Third, we derive the formula for the tangent of a <em>half</em> angle:

tan (θ / 2) = sin (θ / 2) / cos (θ / 2)

tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) The formulae for the conversion of coordinates in <em>rectangular</em> form to <em>polar</em> form are obtained by <em>trigonometric</em> functions:

(x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) Let be the point (x, y) = (2, 3), the coordinates in <em>polar</em> form are:

r = √(2² + 3²)

r = √13

θ = atan(3 / 2)

θ ≈ 56.309°

The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°).

Let be the point (r, θ) = (4, 30°), the coordinates in <em>rectangular</em> form are:

(x, y) = (4 · cos 30°, 4 · sin 30°)

(x, y) = (2√3, 2)

The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) Let be the <em>linear</em> function y = 5 · x - 8, we proceed to use the following <em>substitution</em> formulas: x = r · cos θ, y = r · sin θ

r · sin θ = 5 · r · cos θ - 8

r · sin θ - 5 · r · cos θ = - 8

r · (sin θ - 5 · cos θ) = - 8

r = - 8 / (sin θ - 5 · cos θ)

The <em>linear</em> function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

To learn more on trigonometric expressions: brainly.com/question/14746686

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