(87) Evaluating Inverse Trigonometric Functions

Now let's talk about evaluating inverse sine functions. So let's say if we want to find the value of the inverse sine of one half, what is that equal to? What do you think the answer is? Now, sine of what angle is equal to one-half? If we could figure this out, then we could find out what the value of arc sine one-half is.

Now we know sine pi over six, which is the same as 30 degrees, is equal to one half. But also, sine five pi over six is equal to one half as well. So then which one do we use? Does the inverse sine of one half equal to pi over six, five pi over six, or is it both? It turns out that the arc sine of one half is equal to pi over six, but it does not equal five pi over six.

Now why is that? The answer is due to the range of the arc sine function. The range is from negative pi over two to pi over two, which we mentioned before, and five pi over six is not in this range. That's why five pi over six is not the answer. Pi over six is between negative 90 and 90. That is the answer.

So, going back to the unit circle, you need to know that the arc sine function exists in quadrants one and four. So if this value is positive, it's going to be in quadrant one because sine is positive in quadrant one. If we had a negative value, we need to use the answer in quadrant four, and we have to pick an answer that's within the range from negative pi over two to pi over two. So even though quadrant four is between 270 and 360, that is not in this range. You need to use negative 90 to 90.

So keep that in mind. Also, you can confirm this with your calculator. If you type arc sine of one-half, and if it's in degrees, you should get 30 degrees. You won't get 150, which is five pi over six. If you put in radians, then you should get pi over six, or maybe the decimal form of pi over six. So arc sine does not exist in quadrants two and three. If you have an answer in these quadrants, you will get it wrong.

Now let's work on some other examples. What is arc sine of the square root of three over two? So go ahead and find the answer. Now we know that sine of, let's use degrees, sine of 60 degrees is equal to the square root of three over two. Also, sine of 300 is also root three over two. And at the same time, I take that back; sine of 300 is negative square root three over two. Sine of 120 is positive root three over two.

So we have two possible answers: 60 and 120. So which one should we use? Now 60 is in quadrant one, which is okay. 120 is in quadrant two, and the arc sine function does not exist in quadrant two. So the answer is 60 degrees, which we can write as pi over three. In mind, pi is 180; 180 divided by 3 is 60. So that's the answer for this example: arc sine of root 3 over 2 is 60.

Now what about arc sine of negative one-half? What's the answer for that? So sine of what angles is equal to negative one-half? Let's make a list and then let's eliminate the wrong answers. So we know sine 30 is one half, and sine 150 is one half. Sine 210 is negative one half; 210 and 150, they all have a reference angle of 30. And 210 is in quadrant three. Sine of 330, which is in quadrant four, that's also negative one half. And also sine of negative 30, that's in quadrant four as well; that's negative one half.

So which one is the correct answer? So we know that the inverse sine function exists in quadrants one and four. It does not exist in two and three. So 210 is in quadrant three, so that answer is eliminated. Now 330 is in quadrant four, and negative 30 is also in quadrant four. So this is 330, and this is negative 30. So which one is the answer? Because they both exist in the appropriate quadrant.

Now don't forget about the range of the arc sine function. It's from negative pi over two to pi over two, or negative 90 to positive 90. 330 is not in that range. 330 is the same as eleven pi over six. So your answer has to be between negative 90 degrees and 90 degrees. So this answer is within the appropriate range, and it's in the right quadrant. Therefore, sine arc sine of negative one-half is negative 30, or negative pi over six. That's the answer that we want.

Now let's work on another example. Go ahead and evaluate arc sine of negative square root 2 over 2. So sine of what angles is equal to negative root 2 over 2? We know that sine 45 is equal to the square root of 2 over 2, but it's a positive answer, not a negative answer. Sine of 225 is equal to negative root 2 over 2, and sine of 315 is also equal to that. And the coterminal angle of 315, which is negative 45, sine negative 45 is negative root 2 over 2.

Now we're going to eliminate this answer because that's in quadrant three, and the inverse sine function does not exist there. And we're going to eliminate this answer because it's not between negative 90 and 90. So this is the only answer that is within the range of negative 90 to 90 degrees, and it's in quadrant four. So arc sine of negative square root two over two is therefore equal to negative 45 degrees, which is the same as negative pi over four.

Now what about these two? What is arc sine of zero, arc sine of one, and the inverse sine of negative one? Find the values of these three things. Now we know that sine of zero is equal to zero, so this answer is zero degrees. Zero degrees is within the range of negative pi over two to pi over two. Now arc sine of one is pi over two, so that's also in the range of the arc sine function. Arc sine of negative one is negative pi over two.

And so those are some other values that you want to be familiar with.

Now let's move on to evaluating inverse cosine functions. So let's evaluate inverse cosine one half. So what's the answer for this problem? Find the exact value of this function. Well, we know that cosine of pi over three, that's cosine of sixty degrees, is equal to one-half. Cosine of five pi over three, which is in quadrant four, is also equal to one-half. Cosine is positive in quadrants one and four but negative in quadrants two and three.

So which one is the answer? Is it pi over three or five pi over three? Now recall that the range of a cosine function, I mean of an arc cosine function, is from zero to pi. So therefore, arc cosine only exists in quadrants one and two. So if you have a positive value like what we do in this case, like what we have here, you want to make sure your angle is between 0 and 90. If it's negative, it has to be between 90, which is pi over 2, and 180, which is pi.

So cosine or arc cosine does not exist in quadrants three and four. So just keep that in mind. Therefore, we can't use five pi over three; that's in quadrant four. So arc cosine one half is equal to pi over three, which is the same as 60 degrees, and so that's the answer.

Now what about this one? What is the inverse cosine of negative square root 3 divided by 2? So take a minute and try that problem. Cosine of 30 degrees is positive square root 3 over 2, but we need a negative value. Cosine is negative in quadrants two and three. Cosine one fifty is negative root three over two, and cosine two ten is also negative root three over two.

We can eliminate 30 because it's not negative; it's positive. And we can eliminate 210 because it's not in the range of zero to pi. 210 is in quadrant three, and the arc cosine does not exist in quadrant three. 150, on the other hand, is in quadrant two, and the arc cosine function exists in quadrants one and two. And also, 150 is in the range; it's between 0 and 180. So 150 is the answer, which is the same as 5 pi over 6. So arc cosine negative root 3 over 2 is equal to 5 pi over 6.

Go ahead and try this one: arc cosine negative square root two over two. So if you have a negative value, you need to choose the angle in quadrant two. Now we know that cosine of 45 degrees is positive square root two over two, but it's not negative. So this is the reference angle that we need. In quadrant two, the angle is 135, which has the same reference angle as 45, and this is negative square root 2 over 2.

So therefore, arc cosine negative root 2 over 2 is 135, and this answer is in quadrant two. It's between zero and pi, and this is the same as three pi over four. So that's the answer for this problem. Now go ahead and try these: arc cosine of zero, arc cosine one, and arc cosine negative one. And remember the range is from zero to pi.

Now we know that cosine of pi is equal to negative one. Cosine of pi over two is zero. Cosine of three pi over two is zero, and cosine of zero is one. So arc cosine of zero must be pi over two. Based on this one here, if cosine pi over two is zero, arc cosine of zero is pi over two, and pi over two is within the range, so that's the answer.

Now what about arc cosine of one? Well, cosine zero is one, so therefore arc cosine one must be equal to zero, and zero is included in the range, so that answer is acceptable. Three pi over two is not between zero and pi. Three pi over two is 270 degrees; it's not between zero and one, so we can get rid of that one. Now pi is in the range. If cosine pi is equal to negative one, arc cosine of negative one is pi.

So these are the answers.

Now let's talk about how to find the exact value of our tangent functions. So let's start with the arc tangent of zero. Tangent of what angle is equal to zero? How can we figure this out? It turns out that tangent of zero degrees is equal to zero. So here's how we know that. We know that tangent of an angle is equal to y over x, and at zero degrees, we have the point (1, 0). So y is zero, x is one; zero divided by one is zero. So tangent of zero degrees has a value of zero.

So this is the angle, and this is the value. When dealing with inverse trig functions, you have the value on the inside, and you're looking for the angle. So the angle is zero degrees.

Now what about this one? What is the inverse tangent of one? What's the answer for that? So tangent of what angle is equal to one? This occurs when y and x have the same value. At an angle of 45 degrees, we have the point (square root 2 over 2, square root 2 over 2).

So tangent of 45 is equal to y over x, and whenever you divide two numbers that are the same, they will equal to one. So arc tangent of one is 45 degrees, which is the same as pi over four. That's the answer.

Now if arc tangent of 1 is equal to pi over 4, what is the value of arc tangent of negative 1?

Now you need to know that the inverse tangent function has a range that's similar to the inverse sine function, and that is from negative pi over 2 to pi over 2. So if we were to draw the unit circle, the inverse tangent function, like the inverse sine function, exists only in quadrants one and four, between negative pi over two and pi over two. The inverse cosine function exists in quadrants one and two; that is between zero and pi.

So the only answer that's going to give us negative one is at 45 degrees; that is negative 45 in quadrant four. So this has to be negative pi over 4, which is in this region right here. We can't use 7 pi over 4 because that's not in this range, so we have to use negative pi over 4.

Now what is the value of the inverse tangent of the square root of three? So because we have a positive value, the answer has to be in quadrant one. So let's use the 30-60-90 reference triangle for this example. Across the 30 is 1, across the 60 is root 3, across the 90 is 2. We know tangent is opposite over adjacent. If we choose the 30-degree angle, opposite is 1, adjacent to it is root three. One over root three is not the same as the square root of three, so therefore the angle that we have to use is 60.

Opposite to 60 is the square root of 3, adjacent to 60 is 1. So tangent of 60 degrees is equal to the square root of 3. So arc tangent of square root 3 is 60 degrees, or pi over 3, which is in quadrant one. If you get an answer in quadrant one, it's okay because arc tangent, arc sine, and arc cosine can all exist in quadrant one.

Now what about the arc tangent of negative square root three divided by three? What is that equal to? So tangent of what angle is equal to negative root 3 over 3? Now if tangent 60 is the square root of 3, we know tangent 30 has to be root 3 over 3. So 30 is the reference angle, but we need it to be in quadrant 4. So it's going to be negative 30.

We can confirm it. Tangent of negative 30, which is negative pi over 6, is equal to sine of negative pi over 6 or negative 30 divided by cosine of negative 30. Sine negative 30 is negative one-half; sine is negative in quadrant four, but cosine is positive. Cosine negative 30 is root three over two. If we multiply the top and bottom by two, we can get rid of these, and so we're left with negative one over the square root of three.

Now we have to rationalize it, so this becomes a negative square root of three over three. Therefore, tangent of negative pi over six is equal to negative square root three over three. So the arc tangent of negative root three over three has to be negative pi over six. This exists in quadrant four, and it's within a range of negative pi over two to pi over two, so that's the answer.

So now you know how to evaluate inverse trigonometric functions.

So to review, remember this: arc cosine exists in quadrants one and two, so that's with a range of 0 to pi. So as long as your answer is between 0 and pi, or within quadrant 1 and 2, you should be okay when evaluating arc cosine functions. The arc cosine function does not exist in quadrants three or four, so I'm just going to put an x in blue.

Now for arc sine, which I'm going to use red, arc sine exists in quadrants one and quadrant four, so that's within a range of negative pi over two to pi over two. So arc sine does not exist in quadrants two and three. And then finally, arc tangent is the same for arc sine. So arc tangent exists in quadrants four and one, and that is between negative pi over two and pi over two. It does not exist in two and three; none of them exist in quadrant three, so you should never use an angle there.

So remember, arc cosine has a range of zero to pi, so your answer has to be in that range. For arc sine and arc tangent, when getting an answer, it has to be between negative pi over two and pi over two.

Now let's say if we have a composition of an inverse trig function, what should we do?

So for example, let's say if we want to evaluate arc sine of sine pi over three, what's the answer for this particular example? It turns out that arc sine and sine will cancel, giving us an answer of pi over three. And we can do that because pi over three is in quadrant one, and arc sine exists in quadrant one. But let's take it one step at a time.

So first, let's evaluate sine pi over 3, or sine 60 degrees. So using the unit circle or using the 30-60-90 reference triangle, you know that sine of 60 is equal to the opposite side, which is square root 3, divided by the hypotenuse 2. So sine 60, or sine pi over 3, is the square root of 3 over 2.

Now we need to evaluate arc sine. Arc sine exists in quadrants one and four. Now because this is positive, we don't want the answer in quadrant four because arc sine is negative in quadrant four. So therefore, the angle has to be 60. It's going to be pi over 3 because that is in quadrant 1, and arc sine exists there. So in this case, the final answer is simply pi over 3. We were able to cancel these two.

The notes provide a detailed examination of evaluating inverse trigonometric functions, specifically focusing on inverse sine, cosine, and tangent functions. Here is a breakdown of the content:

Inverse Sine Function (arcsin)

1. Definition: The inverse sine function, denoted as arcsin, finds the angle whose sine is a given value.

2. Key Example: To find arcsin(1/2), we determine the angle whose sine equals 1/2:

   • We know that sine(pi/6) or sine(30 degrees) equals 1/2.

   • Another answer could be sine(5pi/6), but arcsin only returns pi/6 because the range of arcsin is from -pi/2 to pi/2. Thus, arcsin(1/2) = pi/6.

3. Quadrants: The arcsin function exists in quadrants one and four, meaning:

   • If a sine value is positive, the result will be in quadrant one (0 to 90 degrees).

   • If negative, it would be in quadrant four (-90 to 0 degrees).

4. Range: The results must stay within the range of -pi/2 to pi/2, eliminating any angles outside this interval (e.g., 5pi/6).

5. Additional Examples: Similar evaluations for arcsin(sqrt(3)/2) yield pi/3 (60 degrees) because it is the legitimate angle within the necessary range.

   • For arcsin(-1/2), the angle is -pi/6 (-30 degrees) as it is in the fourth quadrant.

Inverse Cosine Function (arccos)

1. Definition: The inverse cosine function, denoted as arccos, identifies the angle whose cosine is the given value.

2. Example: When computing arccos(1/2):

   • We know cosine(pi/3) (60 degrees) equals 1/2 but are also aware of cosine(5pi/3) as another potential answer.

   • However, since the range for arccos is from 0 to pi, arcsin only returns pi/3.

3. Quadrants & Existence: Arccos only exists in quadrants one and two, limiting possible angle outputs between 0 and 180 degrees (0 to pi). Therefore, results in quadrants three (e.g., 5pi/3) are invalid.

4. Additional Examples: Evaluating arccos(-sqrt(3)/2) gives us 150 degrees (5pi/6), as it is within the accepted range. Similarly, arccos(-sqrt(2)/2) yields 135 degrees (3pi/4).

Inverse Tangent Function (arctan)

1. Definition: The inverse tangent function, denoted as arctan, identifies the angle whose tangent is a provided value.

2. Example: For arctan(0):

   • The angle that gives a tangent of 0 is clearly 0 degrees.

3. Additional Evaluations: For arctan(1), since tangent(45 degrees) is equal to 1; thus, arctan(1) would be pi/4 (45 degrees) and arctan(-1) results in -pi/4 (quadrant four).

4. Range & Existence: The range for arctan is similar to arcsin, being -pi/2 to pi/2, confirming that values must remain within this limit.

Summary of Ranges and Quadrants

• arc sine: Exists in quadrants one and four; range from -pi/2 to pi/2.

• arc cosine: Exists in quadrants one and two; range from 0 to pi.

• arc tangent: Exists in quadrants one and four; range from -pi/2 to pi/2.

Composite Functions

• When combining functions, for example, evaluating arcsin(sin(pi/3)), we conclude the result remains pi/3, as this falls within the valid quadrants/ranges. Thus, arcsin and sin cancel each other out when within their appropriate intervals.