Question

Suppose that the terminal point determined by t is the point 4 /7 , √33 /7 on the unit circle. find the terminal point determined by each of the following.

−t
(x, y) =______

Answer 1

Certainly! Here are the terminal points for each operation:

(a)
`\(t (x, y) = \left((4)/(7), (√(33))/(7)\right)\)`

(b)
`\(4 + t (x, y) = \left((32)/(7), (28 + √(33))/(7)\right)\)`

(c)
`\(-t (x, y) = \left(-(4)/(7), -(√(33))/(7)\right)\)`

(d)
`\(\pi (x, y) = \left((4\pi)/(7), (\pi√(33))/(7)\right)\)`

Given:
`\(t = \left((4)/(7), (√(33))/(7)\right)\)` on the unit circle.

(a)
`\(t (x, y) =\)` The terminal point
`\(t\)` is already given as
`\(\left((4)/(7), (√(33))/(7)\right)\).`

(b)
`\(4 + t (x, y) =\)` To find this, add 4 to both the
`\(x\)` and
`\(y\)` coordinates of
`\(t\)` :

Given
`\(t = \left((4)/(7), (√(33))/(7)\right)\)`

Add 4 to both
`\(x\)` and
`\(y\)` :

`\[4 + t = \left(4 + (4)/(7), 4 + (√(33))/(7)\right)\]`

`\[4 + t = \left((28)/(7) + (4)/(7), (28)/(7) + (√(33))/(7)\right)\]`

`\[4 + t = \left((32)/(7), (28 + √(33))/(7)\right)\]`

(c)
`\(-t (x, y) =\)`To find the negative of
`\(t\)`, negate both the
`\(x\)` and
`\(y\)` coordinates:

Given
`\(t = \left((4)/(7), (√(33))/(7)\right)\)`

Negate both
`\(x\)` and
`\(y\)`:

`\[-t = \left(-(4)/(7), -(√(33))/(7)\right)\]`

(d)
`\(\pi (x, y) =\)`Multiplying
`\(t\)` by
`\(\pi\)` involves multiplying both coordinates by
`\(\pi\)`:

Given
`\(t = \left((4)/(7), (√(33))/(7)\right)\)`

Multiply both coordinates by
`\(\pi\)`:

`\[\pi t = \left(\pi \cdot (4)/(7), \pi \cdot (√(33))/(7)\right)\]`

`\[\pi t = \left((4\pi)/(7), (\pi√(33))/(7)\right)\]`

Therefore, the terminal points for each operation are:

(a)
`\(t (x, y) = \left((4)/(7), (√(33))/(7)\right)\)`

(b)
`\(4 + t (x, y) = \left((32)/(7), (28 + √(33))/(7)\right)\)`

(c)
`\(-t (x, y) = \left(-(4)/(7), -(√(33))/(7)\right)\)`

(d)
`\(\pi (x, y) = \left((4\pi)/(7), (\pi√(33))/(7)\right)\)`