### Video Transcript

In this video, we’ll learn how to
use the Pythagorean identity and double-angle and half-angle formulas to evaluate
trigonometric values. Our first task is to derive these
double-angle and half-angle identities for the sine, cosine, and tangent functions
based on what we already know.

When it comes to trigonometric
relationships we’re already familiar with, we can recall that given some angle 𝑥,
where 𝑥 may be in degrees or radians, the sine squared of that angle plus the
cosine squared of that angle equals one. This is known as the Pythagorean
identity. Along with this, we can recall a
set of equations called the sum and difference formulas. These describe relationships for
the sine, cosine, and tangent functions of two angles, we’ve called them 𝛼 and 𝛽,
that are either being added together or subtracted.

Working from these two pieces of
information, we can begin to derive double-angle and half-angle identities. We’ll start with the double-angle
identities, and here is our goal. Given some angle we’ll call 𝜃, we
want to derive equations for the sin of two 𝜃, the cos of two 𝜃, and the tan of
two 𝜃. Let’s begin by generating an
expression for the sin of two times 𝜃. Now, one way to write two times 𝜃
is as 𝜃 plus 𝜃. And we write it this way because
this can remind us of the sum formula for the sine function. That formula shows us how to take
the sine of two angles, here they both happen to be 𝜃, added together. The sin of 𝜃 plus 𝜃 equals the
sin of 𝜃 times the cos of 𝜃 plus the sin of 𝜃 times the cos of 𝜃, or two times
the sin of 𝜃 times the cos of 𝜃. Notice that now we have an
expression for the sin of two 𝜃 entirely in terms of the angle 𝜃 itself. We’ll write this down then as our
first double-angle identity.

Next, let’s find a similar
relationship for the cos of two times 𝜃. Once more, we can write two 𝜃 as
𝜃 plus 𝜃 and once again use the sum formula to see that this equals the cos of 𝜃
times the cos of 𝜃 minus the sin of 𝜃 times the sin of 𝜃, in other words cos
squared 𝜃 minus sin squared 𝜃. This is one way to write the
double-angle identity for the cosine function. Notice though that, in this
relationship, we have a cosine squared term and a sine squared term. Anytime we see cosine squared or
sine squared, that can remind us of the Pythagorean identity. This identity means that given a
sine squared term, we can write that like this, or a cosine squared term like
this.

In our equation for the cos of two
𝜃, we can make both of those substitutions separately. First, let’s replace the cos
squared of 𝜃 with one minus the sin squared of 𝜃. That gives us this expression,
which we see we’ve obtained through the Pythagorean identity, which simplifies to
one minus two times the sin squared of 𝜃. We can think of this then as an
alternate form of the double-angle identity for the cosine function.

But then there’s even one more form
yet because going back to this version of our cos of two 𝜃 expression, we can now
replace the sin squared 𝜃 term with one minus the cos squared of 𝜃. That gives us this expression,
which simplifies to two times the cos squared of 𝜃 minus one. This then is the third and final
form of the double-angle identity for the cosine function.

Finally then, we can look to
generate a similar expression for the tan of two 𝜃. The tan of two 𝜃 equals the tan of
𝜃 plus 𝜃. Once again, the sum formula lets us
rewrite the form of this expression. And it simplifies to two times the
tan of 𝜃 over one minus the tan squared of 𝜃.

Recording that result, we’ve now
reached our goal. We have expressions for the sin,
cos, and tan of the angle two 𝜃 in terms of the given angle 𝜃. Knowing all this, let’s move on to
driving the half-angle identities for these functions. Our goal has now changed a bit. Given an angle 𝜃, we now want to
derive equations for the sin of 𝜃 over two, the cos of 𝜃 over two, and the tangent
of that angle. As we do this, we’ll use as our
starting point a result that we just derived, the double-angle formula for the
cosine function. Specifically, as we go about
finding an expression for the sin of 𝜃 over two, we’ll use this version of the
cosine’s double-angle identity. We can write it out just like
this.

At first, something may seem off
here because we’re using the cos of 𝜃 and the sin squared of 𝜃 over two. But note that this angle, what
we’ve called two 𝜃, could be any angle so long as it’s twice as big as the angle
we’ve called 𝜃. For this particular form of the
double-angle identity then, so long as the angle for our sine squared is half the
angle for our cosine operation, the identity holds true. Therefore, we can write that the
cos of 𝜃 equals one minus two times the sin squared of 𝜃 over two. This is great because it means we
simply need to rearrange this expression to solve for sin of 𝜃 over two.

Subtracting one from both sides
gives us this result. And then dividing both sides by
negative two cancels that factor on the right. We can rearrange the left-hand side
of this expression so it equals one minus the cos of 𝜃 all over two. And then if we take the square root
of both sides, we get this result that the sin of 𝜃 over two equals the square root
of one minus the cos of 𝜃 all over two.

Before we write this down as our
answer though, let’s recall the unit circle. If we divide this circle up into
four quadrants, we know that the three functions we’ve been working with — sine,
cosine, and tangent — may have different signs depending on which quadrant our angle
occupies. That’s relevant for our expression
for the sin of 𝜃 over two.

Looking on the right-hand side of
this equation, we can notice that because the cos of 𝜃 never is smaller than
negative one and never is bigger than positive one, this whole numerator, one minus
the cos of 𝜃, will never be negative. And that means the entire
right-hand side of this expression will always be positive or zero, never negative
itself.

On the left-hand side though, we
definitely could have the sign of some angle being negative. For example, say that this angle
here on our unit circle is 𝜃 over two. That angle lies in the third
quadrant, where the sign is negative. So applying this equation we’ve
just derived, we would have a negative left-hand side but a positive right-hand
side. In order to allow for this
possibility, on the right-hand side, we’ll lead with a plus and a minus sign. This ambiguity allows for the fact
that while the square root will never be negative, the result on the left may
be. All that said, we now have an
expression for the half-angle equation of the sine function.

Next, we’ll move on to solving for
the cos of 𝜃 over two. And we’ll start doing this in a
similar way as before. We’ll once again use a double-angle
identity for the cosine function. This time, it will be this
form. We choose this form because if we
replace two 𝜃 with 𝜃 and therefore 𝜃 with 𝜃 over two, that means, buried in this
expression, we have the value we want to solve for, the cos of 𝜃 over two.

Adding one to both sides of this
equation gives us this result. And then dividing both sides by
two, canceling that factor on the right, we can take the square root of both
sides. And we find that the cos of 𝜃 over
two equals the square root of cos 𝜃 plus one all over two. Just like with our sin of 𝜃 over
two identity, we have a situation here where the right-hand side will never be
negative but the left-hand side could be. Once more then, we put in these
plus and minus signs in front of the square root.

With that, we now have a half-angle
identity for the cosine function. And our last task then is to find
something similar for the tangent function. As we do this, we’re actually going
to make use of the half-angle identities we’ve derived so far. Recalling that the tangent of an
angle equals the sine of that angle over the cosine of the same angle, we can write
that the tan of 𝜃 over two equals the sin of 𝜃 over two over the cos of 𝜃 over
two. And that equals this expression on
the right.

In this compound fraction, notice
that we have a one over the square root of two in both numerator and
denominator. We can simplify this expression
then by writing it as plus or minus the square root of one minus the cos of 𝜃 over
the cos of 𝜃 plus one.

At this point, to simplify this
expression further, there are two different pathways we can take. Depending on which one we choose,
we’ll get a different form for the tangent half-angle identity. Path one, we can call it, involves
multiplying numerator and denominator of our right-hand side by the square root of
one minus the cos of 𝜃. Notice that this is identical to
the numerator of our original fraction. What results is this fraction. This yields in our numerator one
minus the cos of 𝜃. And if we multiply out all the
terms in the denominator, we get cos 𝜃 minus cos squared 𝜃 plus one minus the cos
of 𝜃. cos 𝜃 minus the cos of 𝜃 equals
zero. So our denominator simplifies to
the square root of one minus the cos squared of 𝜃.

There’s one last simplification we
can make here. And it depends once again on the
Pythagorean identity. One minus the cosine squared of an
angle equals the sine squared of that same angle. And that means we can replace one
minus the cos squared of 𝜃 with the sin squared of 𝜃. But then, the square root of the
sin squared of 𝜃 comes out simply to the sin of 𝜃.

Notice that we no longer have the
plus and minus sign in front of this fraction. This is deliberate because as we’ve
seen one minus the cosine of any angle is never negative. And along with this, for any angle
𝜃, the sine of that angle will have the same sign always as the tan of the half
angle 𝜃 over two. We can write then this expression
for the tan of our half angle 𝜃 over two.

Now, we mentioned earlier that this
is just one of two ways of writing this half-angle identity. That’s because at this step in the
process, we could’ve multiplied by the denominator of our original fraction rather
than the numerator. If we take this pathway, then that
gives us this fraction. Once again, if we multiply together
the two quantities in parentheses, then we have a positive cos of 𝜃 plus a negative
cos of 𝜃 adding to zero. Once more, using the Pythagorean
identity, we can replace one minus the cos squared of 𝜃 with the sin squared of
𝜃. And the square root of sin squared
of 𝜃 equals simply sin 𝜃. For the same reason as before, we
can also leave off the plus or minus signs in front of this fraction. It will always have the same sign
as the tan of 𝜃 over two.

We’ve now derived expressions for
both the double-angle and half-angle identities for the sine, cosine, and
tangent. Keeping them in mind, let’s get
some practice with these identities through an example.

Which of the following is equal to
the square root of one minus the sin of two 𝑥? (A) The absolute value of the cos
of 𝑥 minus the sin of 𝑥. (B) The cos of 𝑥 minus the sin of
𝑥. (C) The absolute value of the cos
of 𝑥 plus the sin of 𝑥. (D) The cos of 𝑥 plus the sin of
𝑥. (E) The sin of 𝑥 minus the cos of
𝑥.

Okay, so here we’re evaluating this
expression. And we want to see which of these
five answer options it equals. The first thing we can notice is
that we’re taking the sin of two times some angle 𝑥. We can think of this then as the
sine of a double angle, where 𝑥 is that angle. Recalling the double-angle identity
for the sine function, we know that the sin of two 𝑥 equals two times the sin of 𝑥
times the cos of 𝑥. Making this substitution into our
square root gives us this result.

And now let’s consider this factor
of one. By the Pythagorean identity, this
unassuming number one is equal to the sine squared of an angle plus the cosine
squared of that same angle. Making that substitution gives us
this expression. So far, it seems as though we’re
complicating rather than simplifying the expression under our square root. But now that we have these three
terms — sin squared 𝑥, cos squared 𝑥, and two sin 𝑥 cos 𝑥 — say that we can
write them as the quantity cos 𝑥 minus sin 𝑥 squared. We see then that our squaring
operation and then taking the square root will effectively invert one another. So we might expect our final result
to be the cos of 𝑥 minus the sin of 𝑥.

Here though, we have to be careful
to make sure this result is truly equal to our original expression. When we think about the sine
function, we know it has a maximum of one and a minimum of negative one. This means if we think about it for
a moment, that one minus the sin of two 𝑥 will never be negative. Since this is true for our original
expression, it must also be true for our final one. However, it’s certainly possible
for the cos of 𝑥 minus the sin of 𝑥 to be negative. To correct this and make our
expression truly equal to the square root of one minus the sin of two 𝑥, we’ll put
absolute value bars around it. And so this is our final
answer. It’s the absolute value of the cos
of 𝑥 minus the sin of 𝑥 that equals the square root of one minus the sin of two
𝑥.

Let’s look now at another
example.

Which of the following is equal to
the square root of one minus the cos of two 𝑥? (A) The absolute value of the sin
of 𝑥. (B) Two times the absolute value of
the cos of 𝑥. (C) The square root of two times
the absolute value of the cos of 𝑥. (D) Two times the absolute value of
the sin of 𝑥. (E) The square root of two times
the absolute value of the sin of 𝑥.

Okay, we want to see about
converting this given expression to one of the five of our answer options. Thinking along those lines, the
first thing we can notice is that we’re taking the cos of two times some angle
𝑥. This suggests we make use of the
double-angle identity of the cosine function. And in fact, there are three
different forms that this identity takes. We can choose any of them. But notice that if we choose this
third one, then upon making that substitution for cos of two 𝑥 under our square
root, we would have negative one being added to positive one adding up to zero. This would simplify the expression
under the square root. So let’s indeed choose this third
form of the double-angle identity.

When we make this substitution,
indeed we find out that this negative one added to a positive one gives us zero. And multiplying all the sines
through, we get the square root of two times the sin squared of 𝑥. This equals the square root of two
times the square root of the sin squared of 𝑥. And here we have to be careful
because we might be tempted to say that the square root of the sin squared of 𝑥
equals simply the sin of 𝑥. Note though that while the square
root of the sin squared of 𝑥 would never be negative, sin 𝑥 by itself could
be. As we simplify this expression
then, we’ll want to include absolute value bars around the sin of 𝑥. This ensures that no matter what
the value of 𝑥, we’ll never get a negative overall result.

Our final answer then is that it’s
the square root of two times the absolute value of the sin of 𝑥 that’s equal to the
square root of one minus the cos of two 𝑥.

Let’s finish up this lesson now by
summarizing a few key points. In this lesson, we derived
double-angle and half-angle identities for the sine, cosine, and tangent
functions. We saw that the double-angle
identities can be derived from the Pythagorean identity along with the sum and
difference formulas. And then the half-angle identities
can be derived using the double-angle identities as starting points. This is a summary of double-angle
and half-angle identities.