Name: Solution
Due Date: Tuesday, May 19th.
Peer Review Assignment 4
Directions: Treat these assignments as if they were take home test questions as they will be graded by your peers
as such.
1. Short Answer: Choose one trigonometric function and explain why the domain must be restricted in order for it to
have an inverse function. Explicitly state the domain and range of both the function and it’s inverse. Use both the
unit circle and the graph of the function to justify your response.
Solution:
I will only give the sin(θ) and cos(θ) solutions. Come ask me if you would like to know the others or can’t figure
them out based on sine and cosine.
sin(θ):
,
By definition f (θ) = sin(θ) is the y coordinate of the point of intersection of the terminal side of the angle θ
and the unit circle. Also by definition a function is invertible if and only if it is one-to-one or if it passes the
horizontal line test. Clearly we can see based on the graph of f (θ) = sin(θ) that it does not pass the horizontal
line test if we include the entire domain. Similarly if we look at the y coordinate of a point on the unit circle
for a given θ we see that we can find infinitely many angles that give the same y value if we just consider all
π
coterminal angles to θ. It becomes necessary then to restrict the domain of sin(θ) from (−∞, ∞) to [ −π
2 , 2 ].
The range will then remain [−1, 1], but we can see that we are only considering inputs or angles in quadrants
1 and 4. Since every y value is unique in quadrants 1 and 4 and similarly we can see that the graph of sin(θ)
passes the horizontal line test in that range we are free to define arcsin(θ) with respect to the restricted domain.
The domain of arcsin(θ) then becomes [−1, 1] and the range is [− π2 , π2 ].
cos(θ):
The same analysis is given to the graph of cosine except that when we restrict the domain we must restrict it
to [0, π] in order for the graph to pass the horizontal line test and the x coordinates on the unit circle to be
unique over the domain. This results in the domain of arccos(θ) to be [−1, 1] and the range to be [0, π].
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2. Give a counter example to disprove the claim that arccos(cos(θ)) = θ for any real number θ. Draw a picture to
justify your counter example.
Solution:
Observe that if we take θ = −7π
6 we get the following angle plot on the unit circle
√
− 3
Thus we have that arccos(cos( −7π
6 )) = arccos( 2 ). However, arccos(x) will only give us an angle in the range of
7π
5π
[0, π]. Therefore, since − 6 is coterminal to 6 which does indeed lie within [0, π] we have that arccos(cos( −7π
6 )) =
√
− 3
5π
arccos( 2 ) = 6 as the picture indicates.
−7π
Therefore since 5π
6 ̸= 6 we have found a counter example that disproves the claim. This is only one such counter
example among an infinite amount. We need only take an angle outside of [0, π] and obtain a similar result.
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