It’s Professor Dave, let’s solve trigonometric equations.
There are lots of problems we can solve using trigonometry, so let’s see what some examples
might be.
The first thing we could do is find things like angles of elevation.
Let’s say that there is a building that is thirty meters tall, and it casts a shadow
that is sixteen meters long.
What is the angle of elevation of the sun to the nearest degree?
Well, we can set up a diagram, and we quickly see that we have a triangle with both legs
of known length.
Which trig function do we know that relates the sides to this angle?
That would be tangent, which is equal to opposite over adjacent.
In this case, tangent theta equals thirty over sixteen, or fifteen over eight.
To solve for theta, we just need to find the inverse tangent of fifteen over eight.
We can plug that into our calculators, making sure we are in degree mode, and we should
get just a hair under sixty-two degrees.
We can do all kinds of useful things of this nature using trig functions.
If we know how far away some building is, we can instantly know how tall it is just
by measuring the angle of elevation required for your line of sight to coincide with the
top of the building.
If this distance is one hundred meters, and this angle is 55.3 degrees, then we can take
the tangent of the angle, which will be equal to this height over one hundred, then multiply
by a hundred, and we get around one hundred forty-four meters.
So these are indeed powerful techniques.
But beyond this, we need to be able to solve trigonometric equations, meaning we need to
be able to look at equations with trig functions in them, and solve for X.
Say we have four sine X plus one equals two sine X plus two.
First we need to combine like terms, so let’s subtract two sine X, as well as one.
That gives us two sine X equals one.
Divide by two, and sine X equals one half.
Now to solve for X, we can definitely take the inverse sine of one half, or we can picture
the unit circle, and recall that pi over six has a sine value of one half.
But the only problem is that this is not the only solution to this equation.
The sine of five sixths pi also gives one half.
Not only that, but either of these values plus any multiple of two pi will also be a
solution, because starting at one sixth pi, if we go all the way around to get here again,
which would be thirteen sixths pi, that will also be a solution.
Therefore, the way we express this solution will be one sixth pi plus two N pi, and five
sixths pi plus two N pi.
With these, N can be any integer, so if it’s one, this is two pi.
If it’s two, we get four pi.
Whatever it is, it represents some number of times going around the unit circle to get
back to the same spot, and therefore, another solution to the equation.
In this way, just as algebraic equations often have zero, one, or multiple solutions, trigonometric
equations often have multiple or even infinitely many solutions.
Things get a little more complicated when we start manipulating the trig functions.
Say we want to solve tangent three X equals one.
Well, what angle gives us a tangent of one?
That would be quarter pi, because root two over two divided by root two over two, gives
us one.
That means that three X equals quarter pi.
But is it just quarter pi?
If we go to the other side, negative root two over two divided by negative root two
over two also gives us one for the tangent, so five fourths pi works as well.
That means that three X equals quarter pi plus N pi.
Now to solve for X, we divide both sides by three.
That leaves us with pi over twelve plus N pi over three.
Trigonometric equations get much more complicated than this.
We can even see them in quadratic form.
Take something like two cosine squared X minus three cosine X plus one equals zero, and we
are just looking for answers in the interval from zero to two pi.
It may look different than your typical polynomial, but we can factor this into binomials just
like we normally would.
We know we will need two cosine X here and cosine X here, then from there, the factors
of one are either one and one, or negative one and negative one.
Since this cosine term is negative, let’s try the negatives, and that does indeed work
out, since we get minus one cosine here, and minus two cosine there.
Now, as we know, each of these binomials can be set equal to zero, giving us two cosine
X equals one, or cosine X equals one half, and then cosine X equals one.
Well, where does cosine equal one half?
That would be at one third pi and five thirds pi.
Where does cosine equal one?
At zero.
So the solutions are zero, one third pi, and five thirds pi.
There are other situations where we will have to factor in a different way.
Take sine X plus two sine X cosine X, solving for the interval from zero to two pi.
These two terms have sine X in common, so let’s factor it out.
That leaves us with sine X times the quantity one plus two cosine X.
Now, if sine X is zero, this whole side is zero, so zero and pi are solutions, because
the sine of zero and the sine of pi both equal zero.
Then for this one, two cosine X equals negative one, and cosine X equals negative one half.
That means that two thirds pi and four thirds pi are also solutions.
Other strategies involve using trigonometric identities.
If we have an expression with sine squared terms and cosine squared terms, we can change
one of them to the other using this identity.
There are too many identities to do examples for each, and as we have already learned the
identities, we can just remind ourselves that all of these strategies are at our disposal.
I think we get the idea, so let’s check comprehension.