In this discussion, we learn how to evaluate trig functions involving inverse trig functions.
For a quick review of how to evaluate inverse trig functions, you can go to the following link. Otherwise, let us continue.
https://www.mathangel369.com/blog/inverse-cosine-find-the-exact-value
In this example, the inner function is an inverse trig function and the outer function is a trig function.
Key things to consider:
🔹Work with the Inner Function First
🔹Observe the Domain of the Inner Function
🔹Observe the Range of the Inner Function
🔹Draw a Triangle
🔹Use Pythagorean Theorem
🔹Find the Value of the Outer Function
Let’s begin!
Say you have, Y = sin[cos^-1(-4/5)].
Note that there is a set of parentheses inside of a set of brackets.
Let
£ = cos^-1(-4/5).
Then
Y = sin[£].
🔷Work with the Inner Function
The first thing that you want to do is to work from the inside out; so, let us work inside of the brackets first.
Inside the brackets, we have
£ = cos^-1(-4/5).
🔷Observe the Domain of the Inner Function
The domain of inverse cosine is [-1,1].
Since -4/5 is within the interval [-1,1], we then take the cosine of both sides to get
cos(£) = -4/5
Note that, given a triangle, the cosine of an angle is equal to the adjacent side of the angle divided by the hypotenuse.
You can also say that the cosine of an angle is equal to the x value over the radius.
Thus,
cos(£) = Adj/Hyp = x/r = -4/5.
Since x is negative, this tells us that we will go left in the negative direction, drawing our triangle in either the second or third quadrant.
Which quadrant do we choose?
To know this, we must consider the range of inverse cosine.
🔷Observe the Range of the Inner Function
The Range of cos^-1(£) is
[0,pi].
🔷Draw a Triangle
Since the x value is negative and the range restricts us from the angle 0 to the angle pi, we must draw our triangle in the second quadrant.
Now let y be the side opposite to the angel £. We want to solve for y (not to confuse lower case y with uppercase Y in our example).
🔷Use Pythagorean Theorem
x^2 ^ + y^2 = r^2
(-4)^2 + y^2 = (5)^2
16 + y^2 = 25
y^2 = 9
y = 3
We choose the positive value when solving for the length of a side of a triangle;
however, we can adjust the sign accordingly (depending on whether we are going up, down, left, or right).
🔷Find the Value of the Outer Function
Now the outer function is
Y = sin(£).
Note that, given a triangle, the sine of an angle is equal to the opposite side of the angle divided by the hypotenuse.
You can also say that the sine of an angle is equal to the y value over the radius.
Thus,
sin(£) = Opp/Hyp = y/r
Now we know that y= 3 and r = 5.
Thus,
Y = sin(£) = y/r =3/5
We are done!
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🎥Here is a suggested video playlist form MathAngel369:
Evaluating Inverse Trig Functions
Evaluating Inverse Trig Functions
https://www.youtube.com/playlist?list=PLr3YZlmqCWZSb-cLBgwoPG6zV00p4aIA_
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🔑 Practice is the key!
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!
Become an Expert in Inverse Trig Functions