Evaluating Trig Functions w/ Unit Circle Degrees & Radians

Evaluating Trig Functions w/ Unit Circle Degrees & Radians

BAM!!! Mr. Tarrou. Ok, we are going to revisit
that unit circle and start evaluating some trig functions that we can figure out. 16
of the basic trig functions…well there is 6 trig functions and 16 angles and use these
to in cooperation to answer some questions, to set up some trig values and keep them in
exact form. We don’t need to use a calculator and most of your Precalculus and trig teachers
should not be allowing you to use your calculators on these questions. Because, you know, we
don’t want just a decimal you get punched out of a calculator. That is not going to
show us that you know this and understand this unit circle. So any trig or precalculus
question that is based off or uses one of these 16 angles, or some half and double angles
which we will learn those identities, but right now we are just using these 16 angels,
they should be left in exact form. That means no decimal numbers like umm.. Like if the
answer is square root of two, leave it as square root of two. Don’t make it 1.datta
datta datta with all the decimals. We have got… When you set up your unit circle don’t
forget that you count all the way around by thirty degrees and then all the way around
using 45 degrees. Or if you are doing radians, count by pi over 6 and then pi over 4. That
will give you all sixteen angles that you need for the unit circle. Also remember that
besides all the quadrantal angles that we just identify easily as moving to the right
or left one unit or up or down one unit from the origin. We do not need a triangle, nor
can we set up a triangle, for those four angles. The only angles that we can set up triangles
for are the 12 angles, each of which you find three in each quadrant. There is only two
triangles making up this entire unit circle. The 60-30-90 triangle and the 45-45-90 triangle.
That is why you see all these repeating coordinates and why look so familiar. The only difference
is, as we can see in these orange and yellow one, in quadrant both the x and the y are
both positive and in quadrant four where we have rotated 330 degrees or 11pi/6 radians
we are using the same 30-60 triangle. But, since we are down in quadrant four the y value
is negative and the thus the same triangle can set up not just two but four different
angles. Ok, let’s move on to evaluating some of these trig functions. The sine of theta
is y over r. The cosine of theta is x over r. The tangent of theta is y over x….and
so on. I am sure you can read the rest of these. Please note that in a trig function
you put in an amount of rotation, either degrees or radians, and you get out the sides of a
triangle. Ok, so in an angle and out a triangle with your six trig functions. Let’s evaluate
a couple. So let’s say that we have the cosine of pi over 3. We have to remember soon that
pi over 3 is 60 degrees. You will also need to remember that…sorry. I forgot one thing.
We are talking about evaluating these trig functions and these trig functions work out
to be the same regardless of the size of the circle, but I have set up a unit circle. When
you set up a unit circle, your r values are always going to be the value of one. That
is the definition of a unit circle. And so if we are holding the r value to be a value
of one on the unit circle, the cosine is just going to be the x coordinate and the sine
is going to be the y. As such I have identified that here in green. So the cosine of pi/3
or the cosine of 60 degrees is just the x coordinate of one half. Let’s go in the opposite
direction. The cosine of negative pi over three. Well positive rotations go in a counter-clockwise
rotation from zero, the positive x axis. And negative pi over three goes in a clock-wise
direction. Negative pi over three is going to land you down here at negative 60 degrees,
or as you can see what I am pointing to negative pi over three is the same as five pi over
three or 300 degrees. Those are coterminal angles, they are separated by 360 or 2pi.
At any rate, it is still going to be the same value at one half. So are you always going
to get the same answer when you take a trig function of an opposite angle like this. The
answer is no. It is not going to work for sine. The sine of pi over three is the y value
at 60 degrees which is square root of three over two. The sine of negative pi over three,
again that clockwise rotation as opposed to counter-clockwise rotation, is going to be
negative square root of three over two. Right there. Why, because sine is based off of the
y coordinate and when you are down in quadrant four, or quadrants three and four, your y
values are going to be negative and thus your sine is going to be negative. We will be learning
later that the cosine function is an even function. I can show you here. When you plug
in opposite values of x you get the same value of y. If you are not sure what even functions
are you can search my channel, type even and odd function into my search window or just
even function or odd function, and you find some videos for that. When you plug in negative
or opposite values of x, you get the same value for y. With the sine function, you plug
in opposite values of x or theta and you get out opposite values of y. I am not to say
x and y in terms of the unit circle, I am just saying in… I probably should be saying
domain and range. You plug in opposite values of the domain and you get the same value for
the range. You plug in opposite for the domain and you get opposite values for range. The
is odd and this is an even function. Let’s take a look at tangent. The tangent of theta
is y over x. Let’s say…let’s find the tangent of 4pi over three. Ok, 4pi/3. That is one-third,
two-thirds, three-thirds, four-thirds. Anything with a divisor of three is a multiple of 60
degrees. So one sixty, two sixty, three sixty, and four sixty. Tangent is y which is negative
square root of three over two over x which is negative one half. You guys…my students
have seen me do this, when you have a fraction over a fraction the bottom one is going to
flip up. That is because you can’t really divide fractions, you have to multiply by
the reciprocal of the second fraction. Well, the second fraction is the denominator. So,
negative square root of three over two divided by negative one half becomes negative square
root of three over two times negative two over one. The two’s cancel out, the negatives
cancel out and you get a final answer of square root of three. Now ok…that is good. Remember
that I picked an angle that is in quadrant three. Now tangent is y over x. In quadrant
three both your x coordinate and you y coordinate are going to be negative thus your tangent
in quadrant three is going to be positive. Just make sure you get that positive answer.
Let’s pick another angle measure. Let’s find the tangent of 270 degrees. I went from radians
to degrees now for you. We are using all these angles off the unit circle by the way, so
we know if we are looking at degrees or radians. If you are using your calculator, and it would
not be for one of these 16 angles, but if you are using your calculator the only time
you have to worry about what mode your calculator is in degree or radians is when you are touching
one of these three buttons, otherwise it does not make any difference. But if you are using
sine, cosine, or tangent, make sure you are in the proper mode. And for those previous
questions, again if they were not one of the sixteen angles on the unit circle, you would
have to radians and now we would have to be in degree mode. Anyway. The tangent of 270
degrees is y over x. It is negative one over zero… hmmm We can’t divide by zero and that
is going to happen, dividing by zero is a possibility with your four quadrantal angles
0, 90, 180, 270, and 360. You could also go in the opposite. Any of those quadrantal angles
(multiples of 90). So tangent, cosecant, secant, and cotangent, all four of these trig functions
have the possibility of occasionally being undefined. So make sure you pay attention
to that and you know that you cannot divide by zero. When we go to graph these trig functions,
these are going to be where you get your vertical asymptotes but we are not there yet. Sine
and cosine will never become undefined because they have divisors or denominators of r. A
circle cannot have a radius of zero otherwise it would just be a point and we would not
be talking about these trig functions. All right. BAM!!! Let’s do a few in reverse. Let’s
do a few inverse trig functions. But first let’s pick uh…uh… Actually let’s pick
an example first. Let’s say we want to look at the cosine of 135. Ok, well the cosine
of 135 is the x value of negative square root of three over two. Excuse me, that is negative
square root of two over two. What if I said what is the inverse cosine of…. Let’s put
it this way. The cosine of theta is negative square root of two over two. Now we can see
in the previous example the complete equation, the cosine of 135 is the negative square root
of two over two. But what if I left the theta empty. What if wanted to find what angle measure
would give me a cosine value of negative square root of two over two? Well I would apply the
inverse cosine function to both sides of the equation and theta would be the inverse, with
the little negative one, the inverse cosine of negative square root of two over two. Now
in a regular trig function you put in an angle measurement and you get out the sides of a
triangle. With an inverse trig function, and we will do a video about this later really
hitting it hard…right now I am just giving you a preview, The inverse cosine of negative
square root of two over two. See I have the sides of the triangle and I want the angle.
So I need an inverse trig function for that. Looking on the unit circle, what angle gives
me a cosine value of negative square root of two over two. Well there is an x value
of negative square root of two over two, so we are getting the 135 degrees that I started
the question off originally with. But, there is another point that has a cosine value of
negative square root of two over two and that is 225 degrees or 5pi over 4. So, that is
all and dandy but let me just give you another preview of what is coming up. If you were
to put that into your calculator and it was in degree mode, the calculator would only
be able to give you 135 degrees. That is because you calculator and computers, especially calculators,
I am sure there is some kind of computer program that can handle this…but um.. Calculators
can only deal with functions. That means that when you plug in an x, it can only give you
one y. So, be aware that when we start to get to use our calculators to do inverse trig
functions we may not necessarily get the answer that we want…or all of the answers. And
there is more answers to this question. I just gave you the answers that are between
0 and 360 degrees. You know we can spin around this unit circle and start talking about coterminal
angles and say well there is an indefinite number of times that I am going to have an
x value that is negative square root of two over two. So we have to restrict where we
want the answers from so that you don’t just keep writing for the rest of your life trying
to fill up all the answers of the inverse cosine of negative square root of two over
two. Ok… BAM!! Alright. Let’s take a look at one of these reciprocal functions. These
functions are not going to be on your calculator by the way. So if you are asked for a secant,
cosecant, or cotangent you will have to pay special care in how you use your calculator.
Or, if it is one of your sixteen angles of course do it with the unit circle. That is
what we are going to do right now. The secant of negative 7pi over 6. So we have a negative
angle measure so that means clockwise rotation and it is in radians. It has a divisor of
6 so that is multiple of 30 degrees. So, 1, 2, 3, 4, 5, 6, 7pi over 6. If you take negative
7pi over 6 and you add 2pi you will get 5pi over 6. These are coterminal angles. Secant
is.. bababababa… r over x, so it is the reciprocal of the x coordinate. That means
that it is negative two over the square root of three. You know that we are never going
to be happy with radicals in the denominator. So we are going to multiply the top and bottom
by the square root of three. We get negative two square root of three over three. That
is an example of how to one of the reciprocal functions on the unit circle. I am Mr. Tarrou.
That is the unit circle. BAM!!! Go Do Your Homework.

98 Replies to “Evaluating Trig Functions w/ Unit Circle Degrees & Radians”

  1. If we divide a circle by one Radian (180/pi, or 57.29577 degrees) and than draw it on a circle, it divides the circle in 6 angles with that measurements and one smaller angle with 16.2254 degree measurement, which does not fit the definition of a radian. 
    Isn't it important for the radian measurement system to divide the circle in equal parts?
    If not, than why 1 radian is called a unit in radian system? or maybe it is not called a unit.

  2. Great video, indeed (I subscribed)! However, I'm still confused as to how you got sec(-7pi/6) to be 5pi/6. I understand that since it's negative, you do clockwise rotation, and that since the denominator was 6, you said that it's multiples of 30. And that's when I started to become confused. Could you just explain how you knew that it would land on 5pi/6. Sorry for such a silly question, but that's where it's confusing to me.

  3. Also, during class today, our teacher pointed out something interesting of how if we were given, say, -pi/4, what would be the positive form of that? And I certainly had no clue how she got 7pi/4. Because it wasn't a lesson that we did, but something we spoke about for literally 2-3mins. But I'm very interested on how she got that. She also did an example with -7pi/4, and the positive form of that would be 5pi/6. So confusing how she got that.

  4. 3 hours of confusion in my class and I learn material from you in minutes! thanks for your enthusiasm, I'll be sure to follow your videos throughout my studies!! BAM

  5. You are a great teacher. Thanks a lot for taking your personal time to bring these videos. They are of tremendous value and will surely kindle the interest of math in many kids!! 

  6. I can very easily run with it , but these "values" of x and y never did come together in my mind to make a valid point. So I must ask, what's the significance of whatever the unit circle is supposed to prove??

  7. I always watch your Videos. 2.5 hrs in calc, and I learned within mins of your Videos. Thank you for sharing. #Trig Chapter covered! #BAMMM!

  8. Found out I passed trig yesterday…much credit to your videos. Thank ya very much sir. I must buy you that beer now 🙂

  9. Great video.This was the first time I was learning unit circle and 'BAM!' I understood it well.Thank you.

  10. It is amazing that so many math tutors shy away from trig. Thank you for explaining trig in a concise easy to understand manner.

  11. It's sad that I learn more from your videos than from my professor and his tutoring. Oh well thanks a bunch!!

  12. Thank you for this!! So much better than Khan Academy, it's so nice to have a face than some voice in hidden in darkness, keep up the awesome work!

  13. but the inverse cosine of negative root 2 over 2 doesnt equal to 235 degrees because doesnt inverse have restrictions?

  14. Mr.RobBob, the last example for sec (-7pi/6), you said sec = r / x; isn't r = 1? making the final answer square root 3 over 3 ? please let me know if I'm wrong or right

  15. Best Professor ever! I remember when I first discovered your videos when I was in 11th grade, now I'm in college and you are still saving me from Math courses! You're the best Mr. Tarrou! 🙂

  16. Hi professor. Can you please teach me how to answer this question?

    "determine the coordinates of the terminal point of θ that lie on a circle with radius 3 if: θ = 11π/6 ? "

  17. hi sir when simplying radical example 1/√3 when multiplying 1 by √3 is it still going to be 1√3 or just √3? and when you're given a fraction who has the same numerator and denominator but the numerator has a radical example √3/3 does it consider as 1 or no? thank you

  18. I know right. Mr. Tarrou is a great professor. when I need the help. his videos are there. I have been watching his videos for 2 and a half years and counting

  19. Im a senior in high school and i have tried everything to figure out this unit circle and the reciprocal functions! i found your videos and they have been so helpful! Thank You!

  20. This video was clear and concise, I learned in 14 mins what my teacher couldn't teach in 2 hours. Keep up the good work 🙂

  21. You are one of my favorite Teacher, thanks a lot for all the videos and I told my Math Lecturer and my friends about you.

  22. re Odd and Even funct's ; When you graph Sine and Cosine without a delay in the Phase, you will see the visual representation of an Odd function for Sine and vice versa for Cosine. Useful if you go on to do stuff like signal processing and electrical engineering.

  23. Helloooo!!! I’m a loyal student watching you’re videos since I was in 8th grade. I commented on one of your videos from your triangle postulates and now here I am studying trigonometry in your channel and I am already in 11th grade. And I thought it was ridiculous for my professor to not allow us to use calculators, oh well 😂😂😂.

  24. Help me ProfRobBob, you're my only help! lol, okay so in this video and many others I've seen… We're using the 30/60 triangle and the 45/45 triangle… but why these triangles? Can we use other triangle ratios to get different angles? Can you mix any triangle angles ? And can you use more than 2? Can you add another to the 30/60 and 45/45? Why are 30/60 and 45/45 always used?

  25. Sir, teacher says sine is the ratio of perpendicular and hypotenuse in a right angle triangle then, how it is possible to find value of sine for obtuse angles

  26. To whoever students: if you dedicate a month (one all nighter) to learn trig and the unit circle here to maximum capacity, you will slay in like all calc.

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