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!