what is (sqrt 2, -pi/4) in rectangular coordinates?
what is (-sqrt 6, -sqrt 2) in polar coordinates?
r = √(x^2+y^2)
tanθ = y/x
so, (-√6,-√2) = (√(6+2),arctan(-√2/-√6)) = (2√2,7π/6)
To go the other way, just recall that
x = rcosθ
y = rsinθ
To convert from polar coordinates to rectangular coordinates, we use the following formulas:
x = r * cos(theta)
y = r * sin(theta)
where r represents the radius or magnitude of the point, and theta represents the angle from the positive x-axis.
1) Converting (sqrt 2, -pi/4) from polar coordinates to rectangular coordinates:
Given r = sqrt(2) and theta = -pi/4,
x = sqrt(2) * cos(-pi/4)
y = sqrt(2) * sin(-pi/4)
Using the trigonometric identities cos(-x) = cos(x) and sin(-x) = -sin(x), the equations become:
x = sqrt(2) * cos(pi/4)
y = -sqrt(2) * sin(pi/4)
Now, we evaluate the trigonometric functions:
x = sqrt(2) * (1/sqrt(2))
y = -sqrt(2) * (1/sqrt(2))
Simplifying, we get:
x = 1
y = -1
Therefore, in rectangular coordinates, (sqrt 2, -pi/4) is equivalent to the point (1, -1).
2) Converting (-sqrt 6, -sqrt 2) from rectangular coordinates to polar coordinates:
To convert from rectangular coordinates to polar coordinates, we use the following formulas:
r = sqrt(x^2 + y^2)
theta = atan2(y, x)
Given x = -sqrt(6) and y = -sqrt(2),
r = sqrt((-sqrt(6))^2 + (-sqrt(2))^2)
theta = atan2(-sqrt(2), -sqrt(6))
Simplifying,
r = sqrt(6 + 2)
theta = atan2(-sqrt(2), -sqrt(6))
r = sqrt(8)
theta = atan2(-sqrt(2), -sqrt(6))
Since the square root of 8 can be simplified as 2 * sqrt(2), we have:
r = 2 * sqrt(2)
theta = atan2(-sqrt(2), -sqrt(6))
Therefore, in polar coordinates, (-sqrt 6, -sqrt 2) is equivalent to the point (2 * sqrt(2), atan2(-sqrt(2), -sqrt(6))).