Sketch a unit circle and label the terminal points corresponding to t= pi/4, 3pi/4, 5pi/4, 7pi/4.
Include the coordinates of each point.
I have 3 other problems like this and i have no clue how to do them. can someone help please?
Sure, I can help you with that!
To sketch the unit circle, start by drawing a circle with a radius of 1 unit. This represents the unit circle because its radius is equal to 1.
Now, let's label the terminal points corresponding to the given values of t. Remember that t represents the angle measured in radians.
For t = π/4, the terminal point is located in the first quadrant of the unit circle. To find the coordinates of this point, you can use the trigonometric functions cosine (cos) and sine (sin). The x-coordinate is given by cos(t) and the y-coordinate is given by sin(t). In this case, cos(π/4) = √2/2 and sin(π/4) = √2/2. Therefore, the coordinates of the terminal point for t = π/4 are (√2/2, √2/2).
For t = 3π/4, the terminal point is located in the second quadrant of the unit circle. Using the same approach as above, cos(3π/4) = -√2/2 and sin(3π/4) = √2/2. Hence, the coordinates for t = 3π/4 are (-√2/2, √2/2).
For t = 5π/4, the terminal point is in the third quadrant. Here, cos(5π/4) = -√2/2 and sin(5π/4) = -√2/2. Thus, the coordinates for t = 5π/4 are (-√2/2, -√2/2).
Lastly, for t = 7π/4, the terminal point falls in the fourth quadrant. Similarly, we get cos(7π/4) = √2/2 and sin(7π/4) = -√2/2. The coordinates for t = 7π/4 are (√2/2, -√2/2).
Repeat this process for the remaining problems, and you'll be able to label the terminal points and their coordinates on the unit circle successfully. If you have any further questions or need help with the other problems, please let me know!
Sure! I can help you with that. Here is a sketch of a unit circle with the terminal points labeled corresponding to t= pi/4, 3pi/4, 5pi/4, and 7pi/4:
(1, 0)
(sqrt(2)/2 , sqrt(2)/2) (0, 1) (sqrt(2)/2, -sqrt(2)/2)
(0, -1)
(-sqrt(2)/2, -sqrt(2)/2) (-1, 0) (-sqrt(2)/2, sqrt(2)/2)
The coordinates of each point are as follows:
- For t = pi/4: The terminal point is located at (sqrt(2)/2, sqrt(2)/2).
- For t = 3pi/4: The terminal point is located at (-sqrt(2)/2, sqrt(2)/2).
- For t = 5pi/4: The terminal point is located at (-sqrt(2)/2, -sqrt(2)/2).
- For t = 7pi/4: The terminal point is located at (sqrt(2)/2, -sqrt(2)/2).
I hope this helps! If you have any other questions, feel free to ask.
x is at cos (angle)
y is at sin (angle)
so for example if angle is pi/4
x = cos(pi/4) = sqrt(2)/2
y = sin(pi/4) = sqrt(2)/2
so ( sqrt(2)/2 , sqrt(2)/2 )
if 3 pi/4 (45 deg in Quadrant 2)
x = -sqrt(2/2)
y = +sqrt(2)/2 etc