Find the terminal points [tex]P(x, y)[/tex] on the unit circle determined by the given value [tex]t[/tex]
(a) [tex]t=\frac{-4 \pi}{3}[/tex].
(b) [tex]t=\frac{8 \pi}{3}[/tex].
(c) [tex]t=\frac{11 \pi}{6}[/tex].
(d) [tex]t=\frac{3 \pi}{2}[/tex].
Suppose [tex](\frac{5}{13},-\frac{12}{13})[/tex] is the terminal point, on the unit circle, determined by [tex]t[/tex]. Find the terminal point determined by each of the following:
(a) [tex]-t+\frac{\pi}{2}[/tex]
(b) [tex]t+\pi[/tex]
(c) [tex]2 \pi-t[/tex]
(d) [tex]t-\frac{3 \pi}{2}[/tex]
Find the reference number [tex]\bar{t}[/tex] for each of the following values of [tex]t[/tex].
(a) [tex]t=\frac{-11 \pi}{9}[/tex]
(b) [tex]t=\frac{11 \pi}{3}[/tex]
(c) [tex]t=\frac{-5 \pi}{6}[/tex]
(d) [tex]t=\frac{17 \pi}{4}[/tex]
Answer (2)
Find the terminal point for t = 3 − 4 π : P ( x , y ) = ( − 2 1 , 2 3 ) .
Find the terminal point for t = 3 8 π : P ( x , y ) = ( − 2 1 , 2 3 ) .
Find the terminal point for t = 6 11 π : P ( x , y ) = ( 2 3 , − 2 1 ) .
Find the terminal point for t = 2 3 π : P ( x , y ) = ( 0 , − 1 ) .
Given the terminal point ( 13 5 , − 13 12 ) for t , find the terminal point for − t + 2 π : P ( x , y ) = ( − 13 12 , 13 5 ) .
Given the terminal point ( 13 5 , − 13 12 ) for t , find the terminal point for t + π : P ( x , y ) = ( − 13 5 , 13 12 ) .
Given the terminal point ( 13 5 , − 13 12 ) for t , find the terminal point for 2 π − t : P ( x , y ) = ( 13 5 , 13 12 ) .
Given the terminal point ( 13 5 , − 13 12 ) for t , find the terminal point for t − 2 3 π : P ( x , y ) = ( 13 12 , 13 5 ) .
Find the reference number for t = 9 − 11 π : t ˉ = 9 2 π .
Find the reference number for t = 3 11 π : t ˉ = 3 π .
Find the reference number for t = 6 − 5 π : t ˉ = 6 π .
Find the reference number for t = 4 17 π : t ˉ = 4 π .
See above
Explanation
Problem Analysis We are asked to find terminal points on the unit circle for given values of t , find terminal points given a known terminal point and transformations of t , and find reference numbers for given values of t . Let's break this down into smaller parts.
Finding Terminal Points for Given t Values (1.1) We need to find the terminal points P ( x , y ) on the unit circle for the given values of t . Recall that x = cos ( t ) and y = sin ( t ) .
(a) t = 3 − 4 π . We have x = cos ( 3 − 4 π ) = − 2 1 and y = sin ( 3 − 4 π ) = 2 3 . Thus, P ( x , y ) = ( − 2 1 , 2 3 ) .
(b) t = 3 8 π . We have x = cos ( 3 8 π ) = cos ( 3 2 π ) = − 2 1 and y = sin ( 3 8 π ) = sin ( 3 2 π ) = 2 3 . Thus, P ( x , y ) = ( − 2 1 , 2 3 ) .
(c) t = 6 11 π . We have x = cos ( 6 11 π ) = 2 3 and y = sin ( 6 11 π ) = − 2 1 . Thus, P ( x , y ) = ( 2 3 , − 2 1 ) .
(d) t = 2 3 π . We have x = cos ( 2 3 π ) = 0 and y = sin ( 2 3 π ) = − 1 . Thus, P ( x , y ) = ( 0 , − 1 ) .
Finding Terminal Points for Transformed t Values (1.2) Given that ( 13 5 , − 13 12 ) is the terminal point for t , we need to find the terminal points for the following: (a) − t + 2 π . We have x = cos ( − t + 2 π ) = sin ( t ) = − 13 12 and y = sin ( − t + 2 π ) = cos ( t ) = 13 5 . Thus, P ( x , y ) = ( − 13 12 , 13 5 ) .
(b) t + π . We have x = cos ( t + π ) = − cos ( t ) = − 13 5 and y = sin ( t + π ) = − sin ( t ) = − ( − 13 12 ) = 13 12 . Thus, P ( x , y ) = ( − 13 5 , 13 12 ) .
(c) 2 π − t . We have x = cos ( 2 π − t ) = cos ( − t ) = cos ( t ) = 13 5 and y = sin ( 2 π − t ) = sin ( − t ) = − sin ( t ) = − ( − 13 12 ) = 13 12 . Thus, P ( x , y ) = ( 13 5 , 13 12 ) .
(d) t − 2 3 π . We have x = cos ( t − 2 3 π ) = − sin ( t ) = − ( − 13 12 ) = 13 12 and y = sin ( t − 2 3 π ) = cos ( t ) = 13 5 . Thus, P ( x , y ) = ( 13 12 , 13 5 ) .
Finding Reference Numbers (1.3) We need to find the reference number t ˉ for the given values of t .
(a) t = 9 − 11 π . Since − 9 11 π is in the second quadrant, t ˉ = ∣ t + π ∣ = ∣ 9 − 11 π + π ∣ = ∣ 9 − 2 π ∣ = 9 2 π .
(b) t = 3 11 π . Since 3 11 π = 3 5 π + 2 π , t is in the fourth quadrant. t ˉ = ∣2 π − 3 5 π ∣ = ∣ 3 π ∣ = 3 π .
(c) t = 6 − 5 π . Since 6 − 5 π is in the third quadrant, t ˉ = ∣ t + π ∣ = ∣ 6 − 5 π + π ∣ = ∣ 6 π ∣ = 6 π .
(d) t = 4 17 π . Since 4 17 π = 4 π + 4 π , t is in the first quadrant. t ˉ = 4 π .
Examples
Understanding terminal points and reference angles on the unit circle is crucial in many fields. For example, in electrical engineering, alternating current (AC) waveforms are modeled using sinusoidal functions, where the phase angle corresponds to the angle on the unit circle. Knowing the reference angles and terminal points allows engineers to analyze and design AC circuits efficiently. Similarly, in navigation, understanding angles and their reference angles helps in calculating bearings and distances.
The terminal points on the unit circle for the given angles are calculated using cosine and sine functions, resulting in specific coordinates for each angle. Additionally, transformations of a known terminal point yield new coordinates that reflect rotational shifts on the unit circle. Finally, reference angles for the specified values of t are determined based on their positions relative to the closest quadrant boundaries.
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