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## Angle addition identities

# Trig angle addition identities

CCSS.Math:

## Video transcript

i've already made a handful
of videos that covers what I'm going to cover, the
trigonometric identities I'm going to cover in this video. The reason why I'm doing it is
that I'm in need of review myself because I was doing some
calculus problems that required me to know this, and I have
better recording software now so I thought two birds with one
stone, let me rerecord a video and kind of refresh
things in my own mind. So the trig identities that I'm
going to assume that we know because I've already made
videos on them and they're a little bit involved to remember
or to prove, are that the sine of a plus b is equal to the
sine of a times the cosine of b plus the sine of b
times the cosine of a. That's the first one, I assume,
going into this video we know. And then if we wanted to know
the sine of-- well, I'll just write it a little differently. What if I wanted to figure out
the sine of a plus-- I'll write it this way-- minus c? Which is the same thing
as a minus c, right? Well, we could just use this
formula up here to say well, that's equal to the sine of a
times the cosine of minus c plus the sine of minus c
times the cosine of a. And we know, and I guess this
is another assumption that we're going to have to have
going into this video, that the cosine of minus c is equal
to just the cosine of c. That the cosine is
an even function. And you could look at that by
looking at the graph of the cosine function, or even at
the unit circle itself. And that the sine is
an odd function. That the sine of minus
c is actually equal to minus sine of c. So we can use both of that
information to rewrite the second line up here to say that
the sine of a minus c is equal to the sine of a times
the cosine of c. Because cosine of minus
c is the same thing as the cosine of c. Times the cosine of c. And then, minus the sine of c. Instead of writing this,
I could write this. Minus the sine of c
times the cosine of a. So that we kind of pseudo
proved this by knowing this and this ahead of time. Fair enough. And I'm going to use all of
these to kind of prove a bunch of more trig identities
that I'm going to need. So the other trig identity is
that the cosine of a plus b is equal to the cosine of a-- you
don't mix up the cosines and the sines in this situation. Cosine of a times
the sine of b. And this is minus--
well, sorry. I just said you don't mix it
up and then I mixed them up. Times the cosine of b minus
sine of a times the sine of b. Now, if you wanted to know what
the cosine of a minus b is, well, you use these
same properties. Cosine of minus b, that's still
going to be cosine on b. So that's going to be the
cosine of a times the cosine-- cosine of minus b is the
same thing as cosine of b. But here you're going to have
sine of minus b, which is the same thing as the
minus sine of b. And that minus will cancel that
out, so it'll be plus sine of a times the sine of b. So it's a little tricky. When you have a plus sign
here you get a minus there. When you don't minus
sign there, you get a plus sign there. But fair enough. I don't want to dwell on that
too much because we have many more identities to show. So what if I wanted an
identity for let's say, the cosine of 2a? So the cosine of 2a. Well that's just the same thing
as the cosine of a plus a. And then we could use this
formula right up here. If my second a is just my b,
then this is just equal to cosine of a times the cosine
of a minus the sine of a times the sine of a. My b is also an a in this
situation, which I could rewrite as, this is equal to
the cosine squared of a. I just wrote cosine of a times
itself twice or times itself. Minus sine squared of a. This is one I guess
identity already. Cosine of 2a is equal to the
cosine squared of a minus the sine squared of a. Let me box off my identities
that we're showing in this video. So I just showed you that one. What if I'm not satisfied? What if I just want it
in terms of cosines? Well, we could break out
the unit circle definition of our trig functions. This is kind of the most
fundamental identity. The sine squared of a
plus the cosine squared of a is equal to 1. Or you could write that--
let me think of the best way to do this. You could write that the sine
squared of a is equal to 1 minus the cosine
sign squared of a. And then we could take this
and substitute it right here. So we could rewrite this
identity as being equal to the cosine squared of a minus
the sine squared of a. But the sine squared of
a is this right there. So minus-- I'll do it
in a different color. Minus 1 minus cosine
squared of a. That's what I just substituted
for the sine squared of a. And so this is equal to the
cosine squared of a minus 1 plus the cosine squared of a. Which is equal to--
we're just adding. I'll just continue
on the right. We have 1 cosine squared of a
plus another cosine squared of a, so it's 2 cosine
squared of a minus 1. And all of that is
equal to cosine of 2a. Now what if I wanted to get
an identity that gave me what cosine squared of
a is in terms of this? Well we could just
solve for that. If we add 1 to both sides of
this equation, actually, let me write this. This is one of our
other identities. But if we add 1 to both sides
of that equation we get 2 times the cosine squared of a is
equal to cosine of 2a plus 1. And if we divide both sides of
this by 2 we get the cosine squared of a is equal to 1/2--
now we could rearrange these just to do it-- times 1
plus the cosine of 2a. And we're done. And we have another identity. Cosine squared of a, sometimes
it's called the power reduction identity right there. Now what if we wanted
something in terms of the sine squared of a? Well then maybe we could go
back up here and we know from this identity that the sine
squared of a is equal to 1 minus cosine squared of a. Or we could have
gone the other way. We could have subtracted sine
squared of a from both sides and we could have gotten--
let me go down there. If I subtracted sine squared of
a from both sides you could get cosine squared of a is equal
to 1 minus sine squared of a. And then we could go back into
this formula right up here and we could write down-- and I'll
do it in this blue color. We could write down that the
cosine of 2a is equal to-- instead of writing a cosine
squared of a, I'll write this- is equal to 1 minus sine
squared of a minus sine squared of a. So my cosine of 2a is equal to? Let's see. I have a minus sine squared
of a minus another sine squared of a. So I have 1 minus 2
sine squared of a. So here's another identity. Another way to write
my cosine of 2a. We're discovering a lot of ways
to write our cosine of 2a. Now if we wanted to solve for
sine squared of 2a we could add it to both sides
of the equation. So let me do that and I'll
just write it here for the sake of saving space. Let me scroll down
a little bit. So I'm going to go here. If I just add 2 sine squared
of a to both sides of this, I get 2 sine squared of a plus
cosine of 2a is equal to 1. Subtract cosine of
2a from both sides. You get 2 sine squared of a is
equal to 1 minus cosine of 2a. Then you divide both sides of
this by 2 and you get sine squared of a is equal to 1/2
times 1 minus cosine of 2a. And we have our other discovery
I guess we could call it. Our finding. And it's interesting. It's always interesting
to look at the symmetry. Cosine squared-- they're
identical except for you have a plus 2a here for the cosine
squared and you have a minus cosine of 2a here for
the sine squared. So we've already found a
lot of interesting things. Let's see if we can do
anything with the sine of 2a. Let me pick a new color
here that I haven't used. Well, I've pretty much
used all my colors. So if I want to figure out the
sine of 2a, this is equal to the sine of a plus a. Which is equal to the sine of a
times the co-- well, I don't want to make it that thick. Times the cosine of a plus--
and this cosine of a, that's the second a. Actually, you could
view it that way. Plus the sine-- I'm just
using the sine of a plus b. Plus the sine of the
second a times the cosine of the first a. I just wrote the same thing
twice, so this is just people to 2 sine of a, cosine of a. That was a little bit easier. So sine of 2a is equal to that. So that's another result. I know I'm a little bit tired
by playing with all of these sine and cosines. And I was able to get all the
results that I needed for my calculus problem, so hopefully
this was a good review for you because it was a
good review for me. You can write these
things down. You can memorize them if you
want, but the really important take away is to realize that
you really can derive all of these formulas really from
these initial formulas that we just had. And even these, I also have
proofs to show you how to get these from just the basic
definitions of your trig functions.