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If P(t)=(cos t,sin t) has coordinates (0.707,0.707), find the coordinates of

a. P(t+π)
x =
y =
b. P(−t)
x =
y =
c. P(t−π)
x =
y =
c. P(−t−π)
x =
y =

Respuesta :

sqdancefan

Answer:

  a. (-0.707, -0.707)

  b. (0.707, -0.707)

  c. (-0.707, -0.707)

  d. (-0.707, 0.707)

Step-by-step explanation:

Sine and cosine have the characteristic that ...

  sin(-t) = -sin(t) . . . . . sine is an odd function

  sin(t ± π) = -sin(t)

  cos(-t) = cos(t) . . . . .cosine is an even function

  cos(t ± π) = -cos(t)

Then ...

  P(t ± π) = -P(t)

  P(-t) = (cos(t), -sin(t)) = (1, -1)×P(t)

  P(-t ± π) = (-1, 1)×P(t)

Using these relations, we get ...

a. x = -cos(t) = -0.707

   y = -sin(t) = -0.707

__

b. x = cos(t) = 0.707

   y = -sin(t) = -0.707

__

c. x = -cos(t) = -0.707

   y = -sin(t) = -0.707

__

d. x = -cos(t) = -0.707

   y = -(-sin(t)) = 0.707

Coordinates were 2 integers (Cartesian coordinates) or a symbol as well as a number that point to the specific place on the grid termed as a coordinate plane. It has four dimensions and two axes: horizontal and vertical that are the "x and y" axis, and further explanation can be defined as follows:

Given:

[tex]\to \bold{P(t)=(cos t,sin t)}\\\\\to \bold{coordinates=(0.707,0.707)}\\\\[/tex]

To find:

coordinates =?

Solution:

  • Radians clockwise would make a new point on the opposite edge of the ring, negating both the x and y-coordinates. As little more than a result, their revised coordinates are just as follows: (-0.707, -0.707).
  • P(-t) will merely reflect the point all along the x-axis. All this does is negate the "y-coordinate", allowing the new location to be found (0.707, -0.707).
  • counterclockwise radians Similarly, this will simply negate the x and y coordinates, yielding (-0.707, -0.707).
  • radians counter-clockwise Reflecting thru the x-axis yields (0.707, -0.707), and rotating it yields (-0.707, 0.707).

Therefore, the final answer are "(-0.707, -0.707) , (0.707, -0.707),  (-0.707, -0.707), and  (-0.707, 0.707)".

Learn more:

brainly.com/question/13868040

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