Professor Dave here, let’s learn some important formulas for trig functions.
I know that you’ve probably had enough of trig functions for a while, but there are
some important formulas to mention, so let’s knock them all out in one clip.
First, the sum and difference formulas for sine and cosine, which are as follows.
These tell us that if we want to find the sine or cosine of an angle that is the sum
or difference of two other angles, we can find it in terms of the sine and cosine of
those two angles.
There are derivations for these, but they are a bit involved, so we won’t go over
them here, instead we will just learn how to apply them.
For example, let’s say we want to find the exact value for cosine of five twelfths pi.
This is not part of the unit circle, but if we split it up into three twelfths pi and
two twelfths pi, which reduce to quarter pi and a sixth pi, these are indeed angles for
which we know precise sine and cosine values.
So let’s take this formula here and use quarter pi for alpha and a sixth pi for beta.
The cosine of the sum of these angles, which is the five twelfths pi we want to know about,
will be equal to the cosine of quarter pi times the cosine of a sixth pi, minus the
sine of quarter pi times the sine of a sixth pi.
Ideally, we have these values memorized from the unit circle.
If not, go back and check out my tutorial on this subject.
Otherwise, we can just plug them in.
They will be root two over two times root three over two minus root two over two times
one half.
After multiplying that’s root six over four minus root two over four, which combine to
give root six minus root two over four.
I know this isn’t pretty, but it is an exact value, which can be more valuable than approximating
with a calculator.
The other three formulas work exactly the same way, we just take the angle of interest
and express it as the sum or difference of two angles that are more convenient to evaluate.
Also, if we place the sine of a sum over cosine of a sum, as well as the sine of a difference
over the cosine of a difference, and do a little manipulation, we can get the sum and
difference form for tangents as well, which look like this.
Next let’s look at double-angle formulas.
These can be derived if we use the sum formulas and plug in the same angle for alpha and beta.
Looking at cosine first, let’s put a theta in the place of all the alphas and betas.
Alpha plus beta becomes two theta, and that will equal cosine squared theta minus sine
squared theta.
So that’s the double-angle formula for cosine.
If we do the same for sine, we get the sine of two theta equals sine theta cosine theta
plus sine theta cosine theta, so that’s two sine theta cosine theta.
Lastly for tangent, we get tangent two theta equals two tangent theta over one minus tangent
squared theta.
These are easy to use.
Let’s say we know that the sine of an angle lying in quadrant one is three over five.
What will be the sine of twice that angle?
Well if we know two sides of the triangle, we know the third, it must be four.
So, cosine theta must be four over five.
We just plug these into the formula, and we get two times three-fifths times four-fifths,
which is twenty-four twenty-fifths.
We should also note that there are multiple forms of the double-angle formula for cosine,
because cosine squared can be expressed as one minus sine squared, and sine squared can
be expressed as one minus cosine squared.
So here are the three forms for that, one being the original, and two more given the
two transformations we just mentioned.
Now we can actually take these other two forms of the double-angle formula for cosine and
do something interesting.
Starting with cosine two theta equals one minus two sine squared theta, let’s solve
for sine squared theta.
We bring this sine term over here, bring the cosine term over there, divide by two, and
we have something called a power-reducing formula.
It allows us to express sine squared theta in terms of cosine theta.
Let’s do the same thing for this other form.
Bring the one to the other side, divide by two, and we have cosine squared theta in terms
of cosine theta.
Combining them gives us one for tangent squared theta.
These expressions are useful in their own right, especially in calculus, but if we continue
along with them, we can do another interesting thing.
Let’s say that for each of these, we plug in X over two instead of theta.
They will look more or less the same, except that all of the cosine two theta terms become
cosine X.
Then, we can take the square root of both sides, and on the left we just get sine, cosine,
and tangent of X over two.
These are called half-angle formulas.
They allow us to evaluate trig functions for smaller angles in terms of bigger ones with
known values, and these plus or minus signs before the radical just mean that we have
to look at the quadrant we are in to assign positive or negative values.
For example, what is the sine of one twelfth pi?
Well, that’s not on the unit circle, but one sixth pi is, which is twice this angle.
So the sine of one twelfth pi is equal to the square root of one minus the cosine of
a sixth pi over two.
The cosine of a sixth pi is root three over two, so we have one minus root three over
two, over two, inside the radical.
Let’s turn one into two over two, so that we can combine these to get two minus root
three, over two, over two again, which means two minus root three over four.
We can take the four out of the radical, and we are left with the square root of two minus
root three, all over two, which will be positive since we are in quadrant one.
Not the prettiest thing to see a root within a root, but that’s the exact answer.
The last formulas we will look at are the product-to-sum, and sum-to-product.
These are exactly what they sound like.
For product-to-sum formulas, each term on the left is the product of two trig functions
involving two different angles.
On the right we have trig functions involving sums and differences.
As we said, there are also sum-to-product formulas, where the sum or difference of two
trig functions is expressed as a product.
These are tricky to remember, so it’s unlikely you’ll have to memorize them.
We just want to know they exist so that we can use them if we want to.
Say we want to express sine nine X plus sine five X as a product.
We would put their sum, fourteen X, up here, and their difference, four X, over here.
We simplify, and we get two sine seven X cosine two X.
These can sometimes be useful in verifying trickier identities.
I think that’s plenty of formulas for today, so let’s check comprehension.