Welcome to our video on the properties
of the trigonometric functions.
The goals of the video are to use the reciprocal,
quotient and Pythagorean identities to determine
trig function values.
So here we have the reciprocal identities.
Sine theta is equal to one over cosecant theta,
cosine theta is equal to one over secant theta,
and tangent theta is equal to one over cotangent theta,
and so on.
Now we could derive these if we remember, down here in red,
that sine theta is equal to y over r,
cosine theta is equal to x over r,
tangent theta is equal to y over x,
and so on.
Let me go ahead and just show one of these.
If we take a look at sine theta equal
to one over cosecant theta.
Well, cosecant theta is equal to r over y.
And once we have it here, remember our fraction bar
is just a division symbol, so this is the same
as one divided by r over y.
And then we can change this to multiplication
by multiplying by the reciprocal.
So we'd have one times y over r,
which, of course, would give us y over r,
which is equal to sine theta.
So we could derive all of these in a similar way.
However, for the purpose of this video,
we want to use that information in order to
find trig function values.
So on this problem, given cosine theta equals two thirds,
determine secant theta.
Well on the previous screen we learned that secant theta
is equal to one divided by cosine theta.
So in this case we'd have one divided by, cosine theta's
equal to two-thirds, so we'd have one divided by two-thirds,
which is the same one times the reciprocal of three halves.
So secant theta is equal to three halves.
And here's another one.
We're given that tan theta is equal to five.
We want to determine cotangent theta.
So, again, referring back to the previous screen,
cotangent theta is equal to one over tan theta.
Well, tan theta is just five.
So cotangent five is equal to one-fifth.
Pretty straightforward.
Let's go and take a look at one more.
We're given secant theta and we want cosine theta.
So cosine theta is equal to one divided by secant theta,
which is given to us as negative five halves.
Again, this fraction bar just means division,
so one divided by negative five-halves
is the same as one times the reciprocal
of negative two-fifths.
And we have our cosine theta equal to negative two-fifths.
Let's take a look at the quotient identities.
We have sine theta divided by cosine theta is equal to
tan theta, and cosine theta divided by sine theta
is equal to cotangent theta.
Again, let's take a look at one of these.
In terms of x, y's and r's, we know that sine theta
is equal to y over r,
and cosine theta is equal to x over r,
and if we rewrite this as a division problem,
we would have y over r
divided by
x over r.
Written as a multiplication problem, we'd have
y over r times the reciprocal r over x.
The r's simplify,
and we have y over x, which is equal to tan theta.
So this one can be shown in a similar way.
And the Pythagorean identities are listed here.
Sine squared theta plus cosine squared theta equals one.
Tan squared theta plus one equals secant squared theta.
And one plus cotangent squared theta
equals cosecant squared theta.
And I'll go ahead and show the derivation of this first one
in terms of x, y's and r's.
So, again, we know that sine theta would be y over r.
And here we have y over r squared plus cosine squared theta.
Well, cosine is x over r squared.
And the question is, does this equal one?
So let's go ahead and square these.
We'd have y squared over r squared
plus x squared over r squared.
And here we do see that we have a common denominator,
so we can rewrite this as y squared plus x squared
over r squared.
But remember that r squared is equal to x squared
plus y squared.
That's essentially the Pythagorean theorem.
So we can replace y squared plus x squared
by substitution with r squared.
So we have r squared over r squared which is equal to one.
Again, this came from the fact that this
is equal to r squared.
Okay, let's go ahead and see how we're going to use these.
Find sine theta and tan theta, given that cosine theta
is equal to negative square root three over five.
Now there's quite a bit going on here.
Let's take a look at this first given information.
We know that cosine theta
equals negative square root three over five.
Remember that that is the same as x over r.
R is always positive, so what that tells us is,
in this case x is less than zero.
They're also telling us
that sine theta is greater than zero.
Well, sine theta is equal to y over r,
so if we know this is positive,
and we know r is positive, y has to be positive.
So this implies that y is greater than zero.
The reason this is important, this tells us
which quadrant we're in,
and if x is negative, and y is positive,
we would be in quadrant two.
So with this information we could also determine
that tangent theta would have to be negative.
Let's go ahead and use one of the Pythagorean identities
to find sine theta.
Remember on the previous screen we had
sine squared theta plus cosine squared theta
is equal to one.
So here we'd have sine squared theta
plus cosine squared theta, which is given as
negative square root three over five squared must equal one.
Let's go ahead and isolate the sine squared.
Here we'd have three 25ths.
So subtract three 25ths on both sides.
Now in order to find sine theta, we'd take the square root
of both sides.
So we know sine theta is going to be positive,
so we'll only take the principle square root here.
So we have sine theta equals square root of 22
over the square root of 25, which is five.
So that's one of the values we were supposed to find.
Now we need to find tan theta.
We know that tan theta equals sine theta over cosine theta
Sine theta was square root 22 over five.
Cosine theta was given as
negative square root three over five.
Now we need to simplify this.
These fives would simplify, so what are we left with here?
We're left with tan theta equals square root of 22
over square root three, let's move the negative up front,
and we should probably rationalize this.
So tan theta would have a three in the denominator,
and a square root 66 in the numerator,
and it is negative.
Let's take a look at another.
On this problem, we want to find sine theta and cosine theta
given that tan theta is equal to five 12ths,
and theta is in the third quadrant.
If tan theta's in the third quadrant,
that tells us that both the x coordinate
and the y coordinate would be less than zero,
which means that both sine theta and cosine theta
would be negative.
Okay, there's a couple ways of going about doing this.
Let's go ahead and write some of this down.
We know that tan theta equals five 12ths.
Which we know is equal to y over x.
Now we may jump to the conclusion that x is equal to 12
and y is equal to five, but that's not true,
because it's in the third quadrant.
But what we can conclude is that
y would equal negative five, and x would equal negative 12,
so producing five 12ths for tan theta.
But since we're in the third quadrant,
they are going to be both negative.
Next, we can find r, since we know that r is equal to
the square root of x squared plus y squared.
Well x squared would be 144,
y squared would be 25.
This works out very nicely.
The square root of 169 equals 13.
So now we have x, y and r, we can find any trig value.
But here, we're only trying to find sine theta
and cosine theta.
Sine theta equals y over r, or negative five 13ths.
And cosine theta equals x over r, negative 12 13ths.
So I didn't necessarily use one of the identities
that we discussed in this video,
but there are several ways to solve these.
This is the first one that came to my mind
in terms of x, y's and r's.
I hope you found this video helpful.
Thank you for watching.