It’s Professor Dave, let’s verify trig identities.
We know a lot about trig functions now, including a number of identities they are involved in.
This brings us to an area in trigonometry that can be initially quite infurating, but
it becomes incredibly satisfying once you get good at it, and that’s verifying trig
identities.
The way this works is that we can have an equation with a bunch of trig functions on
both sides.
Sometimes this can be very messy.
But we just simplify terms by applying certain identities until both sides of the equation
say the same thing.
Once we have done this, we have verified the identity.
Let’s start with a simple one and you’ll see what I mean.
What if we have cosecant X times tangent X equals secant X.
What can we do to prove that this is true?
All we have to do is start rewriting things in ways that we know is true.
For example, cosecant X is one over sine X. Tangent X is sine X over cosine X.
And secant X is one over cosine X.
Now on the left side, we notice that the two sine X terms cancel, and we are left with
one over cosine X.
That’s what is on the right side as well, so the identity has been verified.
That one was easy, but it won’t always be this simple.
What about something like this, where sine X tangent X plus cosine X equals secant X.
Now let’s not panic, the easiest thing we can always do first is express individual
trig functions as something else.
That means we rewrite tangent as sine over cosine, and secant as one over cosine.
Now the sine X terms multiply to get sine squared X. Here’s the clever part, the part
that can be frustrating when you can’t figure out what to do.
If we multiply this cosine X by cosine over cosine, it becomes cosine squared over cosine.
That may seem arbitrary, but we do this so that we can combine the fractions, giving
us sine squared plus cosine squared over cosine.
The reason this is useful is that we can now use another identity we know, the one that
says sine squared plus cosine squared equals one.
That means we can change this to one, and we have one over cosine.
Done and done.
So we have already outlined a few strategies that we can try.
One is to rewrite individual identities, another is to use this identity, which can allow us
to change sine squared plus cosine squared into one, or it can allow us to change one
minus sine squared into cosine squared, or one minus cosine squared into sine squared.
Keep those in mind as they will come in handy.
And we’ve also seen that we can combine two fractions into one.
Naturally, we could also take one fraction and split it into two, if there is a sum or
difference in the numerator.
Basically, you just start doing things that make sense algebraically, even if you don’t
know why you’re doing them, and more often than not, the solution will eventually present itself.
Let’s try some more examples.
How about one plus tangent squared equals secant squared.
First things first, let’s change tangent squared into sine squared over cosine squared,
and secant squared into one over cosine squared.
Here’s a hint, we have just one term on the right, but two on the left, so what if
we try to combine these two terms somehow?
Well to combine terms you need a common denominator, so let’s express one as cosine squared over
cosine squared.
Now we can add them together, and we get sine squared plus cosine squared over cosine squared.
Well let’s recall that sine squared plus cosine squared equals one, so there we have it.
This relationship is actually a basic identity we can use for other problems, along with
one plus cotangent squared equals cosecant squared.
Now let’s try a couple that are harder than these.
This one has two fractions on the left, one plus sine over cosine, plus cosine over one
plus sine, and that will equal two secant X. Here’s one where there’s a few different
ways to go.
One strategy that may jump out would be to combine these fractions by getting a common
denominator.
That is a totally valid way to go, but I can think of a way that’s a little quicker.
Take this second fraction.
Whenever we have sums or differences on denominators, it’s sometimes a good idea to try to get
rid of those.
If we remember how we used complex conjugates in manipulating complex numbers, we will know
that the best thing to do here will be to multiply this fraction by one minus sine over
one minus sine.
On the bottom we FOIL that out, and we get one minus sine plus sine minus sine squared.
The sine terms cancel, leaving us with one minus sine squared, which equals cosine squared.
Now we can cancel the cosine on the top and one on the bottom, and we have two fractions
with cosine in the denominator.
This means we can add them together, and we get two over cosine, since the sines cancel.
On the right, two secant is the same as two over cosine, and that’s all there is to it.
How about one more, since these are just so fun.
Sine squared plus cosine squared plus cotangent squared, all over one plus tangent squared,
equals cotangent squared.
Now there are a lot of ways we could go, and even if you pick a way that isn’t the shortest
way, you can still get to the answer, like if we were to divide this into three fractions.
It’ll add some steps, but it’s a totally legitimate way to go.
There is one way that’s probably the easiest, and I’ll give you a hint, it will require
all three of these identities involving squared trig functions.
I’m sure you’ve already spotted a few things we can do.
First, let’s take sine squared plus cosine squared up here and make that one.
Now we have one plus cotangent squared, which is equal to cosecant squared.
Then we can turn the denominator into secant squared.
Just like that, we are almost done.
Now we can rewrite cosecant squared as one over sine squared, and secant squared as one
over cosine squared.
We know that dividing by a fraction is the same as multiplying by its reciprocal, so
we bring that up here and flip it.
That end up as cosine squared over sine squared, which we know is equal to cotangent squared.
Once again, there are many ways we could have done this problem, and as long as you just
do things that are algebraically valid, you’ll probably get there eventually.
That includes expressing complex trig functions in terms of sines and cosines, combining fractions,
separating fractions, and using some of these other identities.
Work with one side, work with the other side, work with both sides, whatever you need to do.
Don’t worry, this is one topic where a lot of practice is the best way to go, so to get
some more of that, let’s check comprehension.