For more information, please see full course syllabus of Trigonometry
For more information, please see full course syllabus of Trigonometry
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Computations of Inverse Trigonometric Functions
Main formulas:
Remember where the inverse trigonometric functions are defined (their domains), and where there answers lie (their ranges).
- Arcsine is defined for each x between -1 and 1 (inclusive) and produces an answer between − (π/2) and (π/2) (inclusive).
- Arccosine is defined for each x between -1 and 1 (inclusive) and produces an answer between 0 and π (inclusive).
- Arctangent is defined for each x between −∞ and ∞ (exclusive) and produces an answer between − (π/2) and (π/2) (exclusive).
Example 1:
Find the arcsines of the following common values: − 1, − [√3/2], − [√2/2], − [1/2], 0, [1/2], [√2/2], [√3/2], 1.Example 2:
Identify each of the arcsine, arccosine, and arctangent functions as odd, even, or neither.Example 3:
Find the arccosines of the following common values: − 1, − [√3/2], − [√2/2], − [1/2], 0, [1/2], [√2/2], [√3/2], 1.Example 4:
Find the arctangents of the following common values: −√ 3, − 1, − [√3/3], 0, [√3/3], 1, √3.Example 5:
Find arcsin(sin(− (7π /6))), arccos(cos(4π /3)), and arctan(tan(− (5π /4))).Computations of Inverse Trigonometric Functions
- Plot the point - [(π)/3] on the unit circle. Note: − [(π)/3] = − 60^{°}
- Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of − [(π)/3] between 0 and π because 0 ≤ arccos ≤π
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π
- Plot the point - [(4π)/3] on the unit circle. Note: − [(4π)/3] = − 240^{°}
- Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of − [(4π)/3] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arcsin ≤ [(π)/2]
- Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]
- Plot the point [(5π)/4] on the unit circle. Note: [(5π)/4] = 225^{°}
- Recall that sine is the y value of the point you just plotted. Now find the angle that has the same tangent value of [(5π)/4] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arctan ≤ [(π)/2]
- Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]
- Plot the point [(5π)/4] on the unit circle. Note: [(5π)/4] = 225^{°}
- Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of [5p/4] between 0 and π because 0 ≤ arccos ≤π
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π
- Plot the point − [(5π)/6] on the unit circle. Note: − [(5π)/6] = − 150^{°}
- Recall that tangent is [x/y]. Now find the angle that has the same tangent value of − [(5π)/6] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arctan ≤ [(π)/2]
- Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]
- Plot the point − [(3π)/4] on the unit circle. Note: − [(3π)/4] = − 135^{°}
- Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of − [(3π)/4] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arcsin ≤ [(π)/2]
- Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]
- Plot the point [(7π)/6] on the unit circle. Note: [(7π)/6] = 210^{°}
- Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of [(7π)/6] between 0 and π because 0 ≤ arccos ≤π
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π
- Plot the point [(4π)/3] on the unit circle. Note: [(4π)/3] = 240^{°}
- Recall that sine is the y value of the point you just plotted. Now find the angle that has the same sine value of [(4π)/3] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arcsin ≤ [(π)/2]
- Draw a line across the y - axis to locate a corresponding point that is between − [(π)/2] and [(π)/2]
- Plot the point [(3π)/4] on the unit circle. Note: [3p/4] = 135^{°}
- Recall that tangent is [x/y]. Now find the angle that has the same tangent value of [(3π)/4] between − [(π)/2] and [(π)/2] because − [(π)/2] ≤ arctan ≤ [(π)/2]
- Draw a line diagonally across to locate a corresponding point that is between − [(π)/2] and [(π)/2]
- Plot the point − [(π)/3] on the unit circle. Note: − [(π)/3] = − 60^{°}
- Recall that cosine is the x value of the point you just plotted. Now find the angle that has the same cosine value of − [(π)/3] between 0 and π because 0 ≤ arccos ≤π
- Draw a line across the x - axis to locate a corresponding point that is between 0 and π
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Computations of Inverse Trigonometric Functions
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
- Intro 0:00
- Inverse Trigonometric Function Domains and Ranges 0:31
- Arcsine
- Arccosine
- Arctangent
- Example 1: Arcsines of Common Values 2:44
- Example 2: Odd, Even, or Neither 5:57
- Example 3: Arccosines of Common Values 12:24
- Extra Example 1: Arctangents of Common Values
- Extra Example 2: Arcsin/Arccos/Arctan Values
Trigonometry Online Course
I. Trigonometric Functions | ||
---|---|---|
Angles | 39:05 | |
Sine and Cosine Functions | 43:16 | |
Sine and Cosine Values of Special Angles | 33:05 | |
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D | 52:03 | |
Tangent and Cotangent Functions | 36:04 | |
Secant and Cosecant Functions | 27:18 | |
Inverse Trigonometric Functions | 32:58 | |
Computations of Inverse Trigonometric Functions | 31:08 | |
II. Trigonometric Identities | ||
Pythagorean Identity | 19:11 | |
Identity Tan(squared)x+1=Sec(squared)x | 23:16 | |
Addition and Subtraction Formulas | 52:52 | |
Double Angle Formulas | 29:05 | |
Half-Angle Formulas | 43:55 | |
III. Applications of Trigonometry | ||
Trigonometry in Right Angles | 25:43 | |
Law of Sines | 56:40 | |
Law of Cosines | 49:05 | |
Finding the Area of a Triangle | 27:37 | |
Word Problems and Applications of Trigonometry | 34:25 | |
Vectors | 46:42 | |
IV. Complex Numbers and Polar Coordinates | ||
Polar Coordinates | 1:07:35 | |
Complex Numbers | 35:59 | |
Polar Form of Complex Numbers | 40:43 | |
DeMoivre's Theorem | 57:37 |
Transcription: Computations of Inverse Trigonometric Functions
Ok we are trying some more examples of computing values of the inverse trigonometric functions.0000
Here we have to find the arctangent of the following common values.0006
I will start with my unit circle.0013
I remember that when arctangent I am always looking for values between –pi/2 and pi/2 because arctangent is always between pi /2 and –pi/ 2.0029
I wrote less than or equal to there, but it should really be less than because arctangent never actually hits – pi/2 or pi/2.0050
That is because tangent itself cannot be evaluated on pi/2 or –pi/2 because you are basically trying to divide by 0 there.0059
I’m going to make a list here of x and arctan _{z} 1 and root 3.0071
It is really useful if you can remember some common values of the tangent function.0098
For example tan (pi/0) is 0.0105
Tan (pi/6) is sine of pi/6 divided by cosine of pi/6, which works out to root 3/30113
Tan (pi/4) is sine of 4 divided by cosine.0126
Well in both are 2/2 so it is one.0132
Tan (pi/3) is sin/cosin of pi/ 3, that is root 3.0136
That tells me what the values of tangent are at these common values, pi/6, pi/4, and pi/3.0143
There is another way to remember the values of the tangent functions, which is to draw this line at x=1.0155
The tangent function is actually where these lines hit that line. That is 1, root 3/3, and root 3.0164
So if you remember your common values of tangent, it is easy to figure out the values of arctangent.0178
Let me start from the bottom because that is the easiest.0186
Root 3 here, what angle has tangent root 3?0190
Well, we just figured that out, that is pi/3.0194
Also, we get that from here.0199
Pi/3 is the angle that has tangent root 3.0201
What angle has tangent 1?0206
Well, there it is right there, it is pi/4.0210
What angle has root 3/3? It is pi/6.0214
From getting these common values over here.0220
What angles has tangent 0? Well, that is 0.0223
The negative ones are little trickier but if you just remember that the values in the fourth quadrant down here.0227
They are just the negatives of the same values we had up in the first quadrant.0235
That is –root 3, -1, and –root 3/3.0242
For –root 3, what is an angle that has that as its tangent? it is –pi/3.0252
For -1, what is an angle that has that as its tangent? It is –pi/4.0259
Finally for –root 3/3, what is an angle that has that as its tangent? It is –pi/6.0264
There are really two things that are keys to being able to find our tangents.0274
The first is knowing your tangents of the common values.0278
Again, if you can not remember those, just remember that tangent is sine over cosine.0282
If you remember the sine and cosine of the common values, it all comes back to those triangles.0295
Then you can figure out the tangents.0299
It is good to remember the common values that you do get from tangent namely root 3, 1, and root 3/3.0302
It is good to remember those numbers. But if you can not remember them, you can work them out from sine and cosine.0311
Once you figure those out, then you can figure out the common values of tangent of the negative angles.0316
You have to remember that arctangent always takes values between –pi/2 and pi/2.0324
Looking for angles between –pi/2 and pi/2 that have the tangents that we have been given.0334
When we are given these tangents, we just remember values between –pi/2 and pi/2 that have those particular tangents.0341
On our last example here, we are given various angles.0000
We have to find the arcsine of their sine or the same for cosine and tangent.0005
It is very important that you would be able to graph this on the unit circle.0012
That is really the way you start the problem like this.0015
There is my unit circle and that is 0, pi/2, pi, 3 pi/2, and 2 pi.0021
Now, -7 pi/6. The negative means you are going the downward direction from 0.0040
If you go a little bit past pi and -7 pi/6 puts you right there, -7 pi/6.0046
So, arcsine of sine of -7 pi/60056
The sine of that angle is the y coordinate and it is positive there.0068
So, it is arcsine and that is a 30, 60, 90 triangle.0076
I know the common values there are ½. It is positive because the y coordinate is positive.0083
It is arcsine of 1 ½ but remember that arcsine is always between – pi/2 and pi/2.0088
I want an angle between –pi/2 and pi/2.0105
Whose sign is ½. Well, sign is the y coordinate.0108
There is an angle who is y coordinate is ½ and that angle is pi/6.0117
My x are, there is pi/6.0126
It is an angle between –pi/2 and pi/2 who is sign is ½.0132
Let us try the next one, arcosine of 4 pi/3.0138
Well, I find 4 pi/3, that is between pi and 2 pi.0148
In fact, that is down here.0153
Its cosine is x coordinate, its cosine is x coordinate and there again we have a 30, 60, 90 triangle.0158
Its cosine of 4 pi/3 is ½.0174
But since we are on the left side of the page, it is negative because the x coordinate is negative there.0183
Now I’m trying to find the arcosine of – ½.0189
Remember in arcosine, you will always look for an angle between 0 and pi.0196
I’m looking for an angle between 0 and pi whose cosine is – ½.0200
There it is right there.0209
That is 2 pi/3.0212
It is between 0 and pi, and its cosine is – ½.0215
So, our answer there is 2 pi/3.0219
Finally, we have the arctangent of the tangent of -5 pi/4.0229
Let us figure out where that angle is on the unit circle.0239
It is negative so we are going around in the downward direction from 0.0242
5 pi/4 is just past pi.0246
We go down on the downward direction and we end up over here at 5 pi/4.0250
So, -5 pi/4. I’m sorry that is -5 pi/4. Let me go back carefully.0260
Its tangent is sine/cosine.0270
That is a 45, 45, 90 triangle.0276
Its tangent is sine/cosine.0285
Sine is root 2/2 because it is positive.0288
Cosine is –root 2/2 that simplifies down to -1.0293
We want the arctangent of -1.0299
Now, arctangent always takes values between -pi/2 and pi/2 arctangent.0302
We are looking for an angle between –pi/2 and pi/2 whose tangent is -1.0317
So, I’m looking for an angle whose tangent is -1.0327
There it is at that 45, 45, 90 triangle.0339
There is my angle whose tangent is -1.0344
And that angle is –pi/4.0348
To recap, what really made it possible to this problem were finding sine, cosine, and tangent to start with.0361
The first thing you need to is your common values of sine, cosine, and tangent.0369
You need to be able to graph these things on the unit circle.0374
You need to identify the 30, 60, 90 triangles and the 45, 45, 90 triangles and know the common values.0378
Remember the common values for the 30, 60, 90 triangles are always ½ and root 3/2.0388
The common values for a 45, 45, 90 triangles are root 2/2 and root 2/2.0398
You need to know those common values and identify which triangle you are working with.0414
And figure out what quadrant you are in to figure out whether the sine and cosine are positive or negative.0423
The way you remember that is by all students take calculus.0430
In the first quadrant, they are all positive.0434
In the second quadrant, sine is positive only.0437
In the third quadrant, tangent is positive.0440
And in the fourth quadrant, cosine is positive.0443
You figure out which quadrant all these things are positive or negative and then you can find your sine and cosine.0448
And then what we have to do is find arcsines, arcosines, and arctangents.0456
The key to that was remembering the common values but also remembering these ranges.0462
Arcsine, it has to be between –pi/2 and pi/2.0470
Arcosine, between 0 and pi.0474
And arctangent, between –pi/2 and pi/2.0477
What we are doing in this step for all three problems was we had the sine or the cosine of the tangent.0482
We were trying to find an angle in the appropriate range that had the correct sine, cosine, or tangent.0490
That is where our arcosine, arsine, and arctangent are.0498
We are looking for an angle whose sign, cosine, and arctangent is the value that you are given.0501
We are looking for an angle inside this range.0510
In each case, we have a value and we found angles inside that range that have the right sine, cosine, or tangent.0514
That is what the answers are.0521
That is the end of our lecture on computations of inverse trigonometric functions.0524
These are the trigonometry lectures for www.educator.com.0529
Hi these are the trigonometry lectures for educator.com and today we're talking about computations of inverse trigonometric functions.0000
In the previous lecture, we learned the definitions and we practiced a little bit with arcsin, arccos, and arctan, you might want to review those a little bit before you go through this lecture.0009
I'm assuming now that you know a little bit about the definitions of arcsin, arccos, and arctan, and we'll practice using them and working them out for some common values today.0020
The key thing to remember here is where these functions are defined and what kinds of values you're going to get from them.0032
Arcsin, remember you start out with the number between -1 and 1, and you always get an answer between -π/2 and π/2.0040
It's very helpful if you remember the unit circle there.0052
Arcsin always gives you an angle in the fourth and the first quadrant between -π/2, 0, and π/2.0055
You're looking for angles in that range that have a particular sine.0068
Arccos, also between -1 and 1, produces an answer between 0, and π.0075
Again, it's helpful to draw the unit circle and keep that in mind.0081
There's 0, π/2, and π.0088
You're trying to find angles between 0 and π that have a given cosine.0095
Arctan, you can find the arctan of any number.0101
Again, you're trying to find an angle between -π/2 and 0, and π/2 that has a given tangent.0110
I say exclusive here because you would never actually give an answer of -π/2 or π/2 for arctan because arctan never actually hits those values.0129
If you think of that coming from the other direction, we can't talk about the tangent of π/2 or -π/2, because those involve divisions by 0.0143
When we're talking about arctan, we'll never get -π/2 or π/2 as an answer.0153
Let's practice finding some common values.0161
Here's some common values that we should be able to figure out arc sines of.0166
Let me start by drawing my unit circle.0172
There's -π/2 and 0, and π/2, remember, our answers always going to be in that range.0183
Let me just graph those common values and see what angles they correspond to.0191
I'll make a little chart here.0197
x and arcsin(x), so we got -1, negative root 3 over 2, negative root 2 over 2, -1/2, 0, 1/2, root 2 over 2, root 3 over 2, and 1.0198
Remember, sin(x), sine is the y-coordinate.0222
I'm looking for an angle that has y-coordinate of -1 to start with, so I want to find an angle that has y-coordinate down there at -1, and that's clearly -π/2.0227
That's the answer.0238
Negative root 3 over 2, that's an angle down there, so the angle that has sine of negative root 3 over 2, must be -π/3.0240
Negative root 2 over 2, that's the one right there, so that's a -π/4.0258
-1/2, the y-coordinate -1/2, is right there, that's -π/3.0267
Arcsin(0), what angle has sin(0)?0277
Well, it's 0.0280
What angle has sin(1/2)?0283
Well, what angle has vertical y-coordinate 1/2?0285
That's π/3.0288
Root 2 over 2, that's our 45-degree angle, also known as π/4.0294
Root 3 over 3, that's our 60-degree angle, also known as π/3.0303
Finally, we know that the sin(π/2) is 1, so the arcsin(1)=π/2.0313
In each case, it's a matter of looking at the value and thinking, "Okay, that's my y-coordinate, where am I on the unit circle?"0321
"What angle between -π/2 and π/2 has sine equal to that value?"0332
Of course, if you know your values of sine very well, then it's not too hard to figure out the arcsin function.0339
You really don't need to memorize this.0345
You just need to know your common values for sin(x) very well, and to know when they're positive or negative.0348
Then you can figure out the values for arcsin(x).0354
In our second problem here, we're asked to find which of the arcsin, acrccos, and arctan functions are odd, even or neither.0360
It's really, the key to thinking about this one is probably to think about the graphs and not so much to think about the original definitions of odd or even.0369
Let me write down the important properties to remember here.0377
An odd function has rotational symmetry around the origin.0382
The way I remember that is that 3 is an odd number and x^{3}, y=x^{3} has rotational symmetry around the origin.0406
Even functions have mirror symmetry across the y-axis and the way I remember that is that 2 is an even number, and the graph of y=x^{2} has mirror symmetry across the y-axis.0423
That's how I remember the pictures for odd or even functions.0458
Now, let me draw the graphs of arcsin and arccos, and arctan, and we'll just test them out.0464
Arcsin, remember you take a piece of the sine graph, there's sin(x) or sin(θ).0473
Arcsin, I'll draw this one in blue.0482
It's the reflection of that graph in the y=x line, that's arcsin(x) in blue.0485
Now, if you check that out, that has rotational symmetry around the origin.0497
So, arcsin(x) is odd.0507
Let's take a look at cosine and arccos.0517
Cosine, remember, you got to snip off a piece of cosine graph that will make arccos into a function.0522
There's cos(θ), and now in blue, I'll graph arccos.0530
There's arccos(x) in blue.0546
Now, that graph is neither mirror symmetric across the y-axis nor is it rotationally symmetric around the origin.0552
So, it's not odd or even, which is a little bit surprising, because even though if you remember cos(θ) is even.0565
It turns out that arccos(x) is not odd or even, even though cos(θ) was an even function.0590
Finally, arctan, let me start out by drawing the tangent graph, or at least the piece of it that we're going to snip off.0597
It kind of looks like the graph of y=x^{3}, but it's not the same as the graph of y=x^{3}.0609
One big difference is that tan(x) has asymptotes at π/2 and -π/2, and of course y=x^{3} has no asymptotes at all.0617
What I just graphed here is tan(θ) and then in blue I'll graph arctan(θ).0627
I'm flipping it across the line y=x.0638
It also has asymptotes neither horizontal asymptotes.0642
That blue graph is arctan(x), and if you look at that, that is rotationally symmetric around the origin.0648
So that's also an odd function.0666
This problem is really kind of testing whether you know the graphs of arcsin, arccos, and arctan look like.0675
If you don't remember those, then you go back to sine, cosine and tangent, and you snip off the important pieces of those graphs, and you flip them around y=x to get the graphs of arcsin, arccos, and arctan.0685
Those are the graphs that I have in blue here, arcsin, arccos, and arctan.0700
The other thing that this problem is really testing is whether you remember the graphical characterizations of odd and even functions.0705
If you know that odd functions have rotational symmetry around the origin, even functions have mirror symmetry across the y-axis, it's easy to check these graphs to just look at them and see whether they have the right kind of symmetry.0717
Of course, what you find out is that arcsin(x) has rotational symmetry, arccos(x) doesn't have either one, arctan(x) also has rotational symmetry.0732
For our third example here, we're trying to find arccos of the following list of common values.0746
Again, it's useful to start with a unit circle here.0753
Once you start with a unit circle, remember that with arccos, you're looking for values between 0 and π.0766
Arccos is always between 0 and π.0779
We're looking for angles between 0 and π that have cosines equal to this list of values.0786
I'll make a little chart here.0796
-1, negative root 3 over 2, negative root 2 over 2, -1/2, 0, 1/2, root 2 over 2, root 3 over 2 and 1.0805
Remember, cosine is the x-value.0823
I'm going to draw each one of these values as an x-value, as an x-coordinate, and then I'll see what angle has that particular cosine.0826
So, -1, the x-coordinate of -1 is over here, clearly that's π, that's the angle π.0834
Negative root 3 over 2, I'll draw that as the x-coordinate, and that angle is 5π/6.0845
Negative root 2 over 2, I'll draw that as the x-coordinate, I know that's a 45-degree angle, so that's 3π/4, that's the arccos of negative root 2 over 20863
-1/2, if we draw that as the x-coordinate, that's a 30-60-90 triangle, that's 2π/3.0879
What angle has cos(0)?0894
That means what angle has x-coordinate 0, that's π/2.0896
What angle has cos(1/2)?0903
Again, a 30-60-90 triangle, that must be π/3.0909
Root 2 over 2, that's a 45-degree angle, so that's π/4.0917
Root 3 over 2, that's a 30-60-90 angle again, that's π/6.0927
Finally, what angle has x-coordinate 1?0939
That's just 0.0943
The trick here is remembering your common values of cosine on the unit circle.0947
I know all the common values of cosine on the unit circle very well because I remember my 30-60-90 triangles, and I remember my 45-45-90 triangles.0952
I know which ones are positive and which ones are negative.0965
Finally, I remember that arccos is always between 0 and π.0968
So, I'm looking for angles between 0 and π that have these cosines.0973
These are the angles that have the right cosines and are in the right range.0981
We'll try some more examples of these later.0985
1 answer
Last reply by: Dr. William Murray
Mon Aug 4, 2014 8:19 PM
Post by adnan akram on July 17, 2014
In example 1 how 30 degrees is pi/3 and 60 degrees is also pi/3 why? according what I know 30 degrees is pi/6 and 60 is pi/3
Thank you!!
2 answers
Last reply by: Dr. William Murray
Mon Oct 21, 2013 7:32 PM
Post by Robert Mills on October 15, 2013
Between example 1 & 3, I'm confused as to why the arcsinx mirror eachother, meaning why is -1 arcsinx = -Ï€/2, but 1 arcsinx = Ï€/2? On the unit circle, the angle for 1 arcsinx reads 3Ï€/2. I'm just missing some knowledge here, thank you!
1 answer
Last reply by: Dr. William Murray
Wed Aug 14, 2013 12:25 PM
Post by Kiyoshi Smith on August 8, 2013
In Example 3 you give arcsin(-(2)^(1/2)/2) as 3pi/2 when it should be 3pi/4. I thought I should point this out.
4 answers
Last reply by: Dr. William Murray
Thu Jul 18, 2013 8:26 AM
Post by Mark Mercready on December 22, 2011
In Example 1 you give arcsin(-1/2) as -pi/3 when it should be pi/6. And for arcsin(1/2) you give pi/3 when it is pi/6. Please correct this.
1 answer
Last reply by: Dr. William Murray
Wed Apr 3, 2013 11:34 AM
Post by Cain Blaha on November 28, 2011
how would you solve if the angle you were trying to find was not part of a special triangle?
Ex. sin(arccos(3/4))
1 answer
Last reply by: Dr. William Murray
Wed Apr 3, 2013 11:27 AM
Post by Tanielle Spencer on October 28, 2011
my unit circle is not matching his although i follow his concept.
7 answers
Last reply by: Dr. William Murray
Thu Mar 27, 2014 4:51 PM
Post by Hero Miles on May 15, 2011
Timothy, 7pi/6 is on the unit circle. In order to solve your problem you have to evaluate sin 7pi/6 first. Then evaluate inverse cosine of that:
arcos(sin 7pi/6)= arcos(-1/2) = 2pi/3
Piece of cake.
1 answer
Last reply by: Dr. William Murray
Wed Apr 3, 2013 11:24 AM
Post by Timothy Harris on February 18, 2011
this video is very helpful except, what about values that are no on the unit circle or the problems aren't exactly like what is shown
ex: arccos(sin 7pi/6)