In this discussion, we learn how to evaluate inverse cotangent.
Here is an outline of this discussion:
🔹 Observe the Domain of the Function
🔹Observe the Unit Circle
🔹Observe the Range
🔹Choose The Value
Before we begin, let us discuss a few concepts, then work an example.
◻️◻️◻️Concepts◻️◻️◻️
When working with inverse trig functions, in particularly, inverse cotangent, the equation is given in the form
y = cot^-1(x).
🔷Observe the Domain
Note that the domain of cot^-1(x) is the set of all x such that cot^-1(x) is defined.
In other words, you can think of the domain as all possible “input” values.
cot^-1(x) is defined on the interval:
(-inf, inf).
Thus, we say that the domain of cot^-1(x) is:
(-inf, inf).
Always check to see if the x-value satisfies the domain.
If the x-value does not satisfy the domain, then there is no solution.
If the x-value is within the domain, then take the cotangent of both sides to get:
cot(y) =x
🔷Observe the Unit Circle
Looking at the unit circle, you will see terminal points in the form, (a,b).
Remember that cot(y) takes the value “a/b”, given coordinates (a,b).
We search for all values of y such that
cot(y) = x, where x= a/b.
Next, consider the range of cot^-1(x).
🔷Observing the Range
While the domain of cot^-1(x) is all possible “input” values, the range is all possible “output” values.
The range of cot^-1(x) is in the interval (0, pi), which is the upper region of the unit circle, but not including 0 or pi.
Note, that it is possible to define cot^-1(x) using a different range, but in this discussion, we define it using (0,pi).
If you would like to know more about how the domain and range was discovered, go the this article.
🔷Choosing the Value
You may find that there are multiple values for y; however, we must choose the value for y that satisfies the range.
Now, we work an example.
◻️◻️◻️Example◻️◻️◻️
You are given
y = cot^-1(-sqrt(3)).
🔷Observe the Domain
Note that the domain of cot^-1(x) is
(-inf, inf).
Since x is equal to -sqrt(3), we know that it satisfies the domain.
Since the x-value is within the domain of cot^-1(x), we take the cotangent of both sides to get:
cot(y) = -sqrt(3)
🔷Observe the Unit Circle
Looking at the unit circle, we search for all the values of y such that
cot(y) = -sqrt(3).
Or sometimes I like to ask myself, “Where on the unit circle does cot(y)= -sqrt(3)?”
Remember, we want cot(y) = a/b= -sqrt(3), given coordinates (a,b).
This means to observe the second and fourth quadrant for coordinates
(a,b) = (-sqrt(3)/2), 1/2)
or
(a,b) = (sqrt(3)/2), -1/2)
because
a/b = – sqrt(3)/2 / 1/2 = -sqrt(3)
or
a/b = sqrt(3)/2 / -1/2 = -sqrt(3).
On the unit circle, there are terminal points
(a,b) = (-sqrt(3)/2), 1/2) at 5pi/6
and terminal points
(a,b) = (sqrt(3)/2, -1/2) at 11pi/6.
Thus, when y= 5pi/6 and 11pi/6 we have:
cot(5pi/6)= (-sqrt(3)/2)/(1/2)= -sqrt(3)
cot(11pi/6)= (sqrt(3)/2)/(-1/2)= -sqrt(3)
🔷Observe the Range
Now, we cannot choose any value for y in our ultimate answer.
We must consider the range of cot^-1(x).
The range of cot^-1(x) is (0, pi), which is the upper region of the unit circle, but not including 0 nor pi.
🔷Choosing the Value
Remember to choose the value for y that satisfies the range!
We see that the only value within the range is
5pi/6.
Thus, we choose 5pi/6
Hence,
cot(5pi/6) = -sqrt(3),
which means that
5pi/6 = cot^-1(-sqrt(3))
and we say that
y= 5pi/6
We are done!
Thank you for reading!
If you have any questions, or if this discussion helps, let me know in the comments!
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If you would like more examples, practice problems with the answers, quizzes with the answers, and more regarding inverse trig functions, consider taking my Math Course on inverse trig functions!
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