Trigonometric Functions of Any Angle

Trigonometric Functions of Any Angle


BAM!!! Mr. Tarrou. Let’s start talking about
setting up trig functions for angles even if we don’t know them. Unit circle, we have
got the sixteen angles…we know all those angles. What about setting up trig functions
for angles that we don’t know such as this question? Given a point
on the terminal side
of negative two, negative five, set up all
six trigonometric functions. Ok, so let’s do that. We are talking about angles in standard
position so we are going to be using our sine of theta is y over r, cosine of theta is x
over r, and so on. And we are going to draw that angle in standard position. There we
go. I know that negative two, negative five is in quadrant three, the sketch is just going
to allow me to have a place to put the numbers. I am not going to bother trying to get these
drawn to scale. So negative x and negative y, that is going to be in quadrant three.
So I am going to draw an angle that is quadrant three. Now I am going to set up the reference
triangle and uh…that will allow us to work out our six trigonometric functions. We have
a point that is on the terminal side of negative two negative five. I am also going to put
those values along the x and the y movement just to make sure that I do not forget those
signs when I go to set up my six trig functions. Now I have a right triangle and it is a reference
triangle because it is agains the x axis. It is also marking off the number of degrees
between the terminal side and the x axis. So a little pythagorean theorem work here
and we got two squared plus five squared. I am not worrying about the negatives because
there really is no such thing as a negative distance and the negatives are going to get
squared away anyway. This all equals r squared. That is going to be four plus twenty-five
equals r squared. So r is going to end up being the square root of twenty-nine. We will
not be happy with a square root in the denominator so we will need to rationalize some of our
final answers. Well the sine of theta is y over r, so it is negative five over the square
root of twenty-nine. We will need to rationalize that by multiplying the top and bottom by
the square root of twenty-nine. And we get negative five times the square root of twenty-nine
over twenty-nine. The cosine of theta? The cosine of theta is x over y, so it is negative
two over the square root of twenty-nine. When you rationalize that you get negative two
times the square root of twenty-nine over twenty-nine. Remember that sine is y over
r and cosine is x over r. In quadrant three those x any y values are negative so make
sure these answers are both negative as well. Tangent of theta is y over x, so it will be
negative five over negative two, or five halves. And then the reciprocal of sine is cosecant.
The cosecant of theta is going to be this value flipped so it is going to be negative
square root of twenty-nine over five. We will not have to rationalize that. The reciprocal
of cosine is secant. That flipped is going to be the square root of twenty-nine over
two and it will be negative of course because we are still in quadrant three. The cotangent,
well that is going to be two-fifths. So we have evaluated all six trig functions in exact
form, there are no rounded decimals anywhere. We have evaluated and found the exact value
of six different trig functions without ever actually knowing what this rotation of theta
is. We just had to know a point on the terminal side of the standard position angle. Pretty
cool uh? Alright… BAM! Moving on:) Let’s do another example. Hope you are enjoying
these videos. If you are please spread the word about my now budding math channel. I
am really enjoying making these lessons. What if I just told you that the cosine of some
particular angle is less than zero and the cosecant of that same particular angle is
greater than zero. Could you tell me from just this little bit of information, now we
are not going to set up the six trig functions because I know nothing about the sides of
the reference triangles, all I know is that cosine is negative and cosecant is positive….
From just this information could you tell me what quadrant the angle actually is in?
We are trying to evaluate and work with trig function without the aid of a calculator telling
use what the actual measure of the angle is. Well, where is cosine going to be negative?
You put angle measures into a trig function and out of that trig function one more time
you get the sides of a right triangle. So the cosine of theta is x over r. Cosecant,
that trig function will have an angle put into it, and it will also give us the sides
of a triangle. Cosecant is the reciprocal of sine so it will give us r over y. Now when
we talk about angles in standard position it is impossible for the radius of a circle
to be negative. So we are only concerned about the sign changes of x and y. Well… So where
are the x coordinates negative? The x coordinates are negative in quadrants two and three. For
cosecant to be positive, the y values must be positive. In what quadrants are the y values
positive? Quadrants one and two. So if you have an angles whose cosine is negative and
the cosecant of that same angle is positive, then where is the overlap? The overlap is
going to answer the question, “What quadrant is theta in?” It looks like theta is in quadrant
two. So we are not evaluating a trig function necessarily, but are using their values to
identity this unknown angle… or at least the quadrant that this unknown angle terminates
in. So that is how you use trig functions and their signs. Go to the x and y ratios
and use them to evaluate what quadrant your angle is in. ALRIGHTY THEN! Now I kind of
did this in another video. I want go from cosine of theta, given that values and then
just find all the other five trig functions. I am going to say that the cosine of theta
is equal to negative three-fifths and that angle theta is in quadrant two. Well, when
I am given cosine I am given an x and an r value. So, I am given two sides of the reference
triangle and it is in quadrant two. So theta looks like. There is a rough sketch of an
angle in quadrant two. Our reference triangle again drawing against the x axis is going
to be…. Always draw your reference angles or reference triangles, the same thing, against
the x axis. Cosine is x over r, so negative three and five. Again just putting that sign
in there as a place holder so I don’t forget it later. Pythagorean Theorem three squared
plus something squared equals five squared. Well three squared plus, I am again ignoring
the negative because there is no such thing as a negative distance…that is a directional
sign, y squared equals five squared. When you get done solving that you are going to
find out the y coordinate is four. Now that we have the three sides of our reference triangle,
I can set any trig function that I like and for time purposes I am only going to do one
of these. The sine of theta is equal to y, which is four, over r which is five. And in
quadrant two our y values are positive, so thus our sine function had better be positive
when I am done setting it up. The tangent function in quadrant two, the x’s are negative
and the y’s are positive so the tangent would be negative. Let me just write that down.
The tangent of theta is negative three-fourths…NO IT IS NOT!!! The tangent of theta is negative
four-thirds. Ok In quadrant two your x’s and y’s have different signs so your tangent which
is y over x should have a negative sign. So that is how you set up these little right
triangle drawings and show the rotation for all these trig function problems. That is
what we are studying, right triangles and their rotation about the origin, or a circle
if I were drawing a unit circle. ALL RIGHTY THEN!!! Are you still awake? Are you with
me? Let’s go to the next problem. The tangent of theta, forgot to write that in my notes,
is equal to five-twelves. I am not going to blatantly tell you what quadrant theta is
in, but I will say that the sine of theta is negative. My tangent ratio which is y over
x is positive. The only way you can divide two values that can have different signs,
the only way you can divide y and x and get a positive answer, is if either both are positive
which happens in quadrant one of if they are both negative. Negative five divided by negative
twelve would give you a positive five-twelves. This angle might be in quadrant three. Now
that I am told the sine of theta is negative, and the sine of theta being y over r, where
are my y values negative? Y values are negative in quadrants three and four…they are positive
in two. Well…HEY! There is my overlap so that is how I am going to draw my angle and
reference triangle, in quadrant three. So just a rough sketch of an angle in standard
form in quadrant three. A quick rough sketch of a reference triangle again against the
x axis. Reference angles is an acute angle and it is the number of degrees or radians
between the terminal side and the x axis, so here is your reference triangle. Tangent
is y over x, so y over x. Now I am in quadrant three so both the x movement and the y movement
are negative. Now the Pythagorean Theorem, five squared plus twelve squared equals 169.
The hypotenuse will be thirteen after you take the square root and finish working through
the Pythagorean Theorem. A squared plus b squared equals c squared. Now I have the three
sides and I can set up any trig function I like. Let’s just do one of these. I want to
get in one more example before the end of my time period. Let’s say I want the sine
of theta, which is y over r. The cosine of theta is x over r. And let’s just again hit
this idea of a reference angle. What if I did only rotate ten degrees? What would be
the reference angle? It would be ten degrees. What if I rotated 170 degrees? How many degrees
away from the x axis am I? I am ten degrees away from the x axis so the reference angle
is going to be ten degrees. What if I rotated negative twenty degrees? What would the reference
angle be? Well it is acute, it is the number of degrees between the terminal side and the
x axis, so the reference angle would be twenty degrees. Almost a matching number but a sign
change. Make sure that you remember that a reference angle is just the number of degrees
between the terminal side and the x axis. Sometimes you compare it to zero degrees,
sometimes you compare it to 180 degrees, sometimes you might calculate it from 360 degrees. If
you rotate 340 degrees, your close to the x axis… I am running out of time so I am
getting a little flustered. If you rotate 340 degrees your reference angle will be 360-340=
20 degrees. Those ideas of reference angles and their importance is all over the unit
circle. I am Mr. Tarrou. BAM!!! Thank you for watching. Go Do Your Homework!

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