convert (-sqrt2, -sqrt6) to polar coordinates with
r > 0 and 0 ≤ θ < 2π.
tanθ = -√6/-√2 = √3 , where θ is in quad III
θ = 180°+60° = 240° or 4π/3 radians
r = √(-√2)^2 + (-√6)^2) = √8
(-sqrt2, -sqrt6) <-----> (√8, 4π/3)
Well, well, well. Converting to polar coordinates, eh? Let's embark on this mathematical journey, my friend!
To begin, we need to find the value of r, which represents the distance from the origin (0,0) to the point (-√2, -√6). Since r has to be positive, we'll take the absolute value of the coordinates: (√2, √6).
Now, let's tackle the angle θ. We need to find the angle between the positive x-axis and the line connecting the origin to our point (-√2, -√6).
Get ready for a twist! Since both coordinates are negative, we're in the third quadrant. The angle θ will be π + another angle, which we can call α. To find α, let's do some trigonometry!
The tangent of α will be the ratio of the y-coordinate to the x-coordinate. Therefore, tan(α) = (√6)/(√2) = √(6/2) = √3.
Using our trigonometric wizardry, we find that α is π/3, as tan^-1 (√3) = π/3.
Now we can determine θ! Remember, θ = π + α, so θ = π + π/3 = 4π/3.
Thus, the polar coordinates for (-√2, -√6) are (r, θ) = (√2, 4π/3).
Voilà! We did it!
To convert (-√2, -√6) to polar coordinates, we can use the following formulas:
r = √(x^2 + y^2)
θ = arctan(y / x)
Given that x = -√2 and y = -√6, let's substitute these values into the formulas.
First, let's calculate r:
r = √((-√2)^2 + (-√6)^2)
= √(2 + 6)
= √8
= 2√2
Next, let's calculate θ:
θ = arctan((-√6) / (-√2))
= arctan(√6 / √2)
= arctan(√3)
≈ 1.2490 radians (in the range of 0 ≤ θ < 2π)
Therefore, the polar coordinates of (-√2, -√6) with r > 0 and 0 ≤ θ < 2π are approximately (2√2, 1.2490 radians).
To convert the point (-√2, -√6) to polar coordinates, follow these steps:
Step 1: Find the value of r (the distance from the origin to the point).
To find r, use the formula:
r = √(x^2 + y^2)
In this case, x = -√2 and y = -√6. Substitute these values into the formula:
r = √((-√2)^2 + (-√6)^2)
= √(2 + 6)
= √8
= 2√2
So, the value of r is 2√2.
Step 2: Find the value of θ (the angle that the line connecting the point to the origin makes with the positive x-axis).
To find θ, use the formula:
θ = arctan(y / x)
In this case, x = -√2 and y = -√6. Substitute these values into the formula:
θ = arctan((-√6) / (-√2))
= arctan(√6 / √2)
= arctan(√3)
= π/3
So, the value of θ is π/3.
However, we need to ensure that θ is in the range of 0 ≤ θ < 2π. Since π/3 is already within this range, we don't need to make any adjustments.
Therefore, the polar coordinates of (-√2, -√6) are (r, θ) = (2√2, π/3).