Main content

# Proof of the law of cosines

Sal gives a simple proof of the Law of cosines. Created by Sal Khan.

## Want to join the conversation?

- Could you explain this using actual values instead of just letters? Like say you were given two sides and an angle?(75 votes)
- Refer to "Trigonometry word problems (part 1)"(115 votes)

- How is the sine of theta squared + the cosine of theta squared equal to one?(37 votes)
- sin x = {opposite side / hypotenuse }

sin^2(x) = [ opposite side / hypotenuse ]^2

cos x = {adjacent side / hypotenuse }

cos^2(x) = [ adjacent side / hypotenuse ]^2

sin^2(x) + cos^2(x) = [ opposite side / hypotenuse ]^2 + [ adjacent side / hypotenuse ]^2

= [ (opposite side)^2 + (adjacent side)^2 ] /( hypotenuse )^2

= (hypotenuse)^2 / (hypotenuse)^2

= 1(86 votes)

- Does anyone have a good link for practice problems for the law of cosines (preferably with the answers) ?(15 votes)
- You can make up your own problems! Cut a triangle out of a piece of paper, measure two sides and an angle, and see how closely you can predict the third side, or measure the three sides and see how closely you can predict the measure of any of the angles.(130 votes)

- how do you know when to use the:

the sine rule

cosine rule

tangent rule.(19 votes)- Sine rule: When you have all the angles and a side, to calculate the other sides. (If you use it the other way, you will find two possible values for the angles, as sin( 80º ) = sin( 100º ), for example.)

Cosine rule: When you have the three sides and want to calculate an angle, or when you have two sides and an angle, and want to find the third side.

Tangent rule: When you want to be fancy.(69 votes)

- Why is (at6:35) it that (c-bcosQ)^2 = c^2 - 2cbcosQ + b^2cos^2Q ? I understand that he multiplied it out but I don't see how he got to that. Can someone please write out the logic?(14 votes)
- ( c - bCosQ ) ^ 2

= (c - bCosQ) x (c - bCosQ)

= (c x c) - (c x bCosQ) - (bCosQ x c) + (bCosQ x bCosQ)

= c^2 - cbCosQ - cbCosQ + b^2Cos^2Q

= c^2 - 2cbCosQ + b^2Cos^2Q(14 votes)

- how do you solve for side "a" if the only measurements you have are angle theta, the angle between sides "b" and "a" (the one from which you dropped the altitude) and the length of side "c"?(14 votes)
- In that case you'd want to use the Law of Sines ( http://www.khanacademy.org/video/proof--law-of-sines?playlist=Trigonometry )!(11 votes)

- how do you figure out which sides are a,b, and c?(13 votes)
- To solve for a, just use your known sides (doesn't matter which) for b and c, and the angle between them to be theta. Then just plug-and-chug the formula to find your unknown. If you are given three sides and need to solve for the angle, make sure the side you plug in for a is opposite the angle you want to find.

Note that some forms of the law solve for c using angle gamma or some other arrangement. For whatever form you are using, just dump the known sides on one side and solve for the unknown on the other, or put the side opposite the unknown angle by itself and dump the other sides into the rest of the function.(4 votes)

- Could this be applied along with the law of sines to make a law of tangents?(9 votes)
- Good question! Not exactly, but there is a "law of tangents" that is equivalent to the law of sines.

http://en.wikipedia.org/wiki/Law_of_tangents(13 votes)

- How do you prove sin^2θ + cos^2θ = 1 ?(9 votes)
- There are many ways to do this. Here is a link that shows several of those proofs:

https://proofwiki.org/wiki/Sum_of_Squares_of_Sine_and_Cosine

I personally prefer the exponential formulation proof because it doesn't invoke any trig identities other than knowing what sine and cosine are in terms of e.(7 votes)

- Not a question - but a comment (smile). The "age" comments were cute -- I did take trig 5 years ago - just prior to retiring from the Army and beginning teaching of math. Now at 58 (a long cry after 40) I'm still doing math, and having to relearn Trig. So - there are some of us after age 40 still doing trig - smile(11 votes)
- Good for you - I am the same age, but Math stuck better in my head, so after a long stint working at the patent office, I went back to teaching high school students math, but most if what I have done so far I did not loose, but whenever I have to help a student with Calculus, they have to help remind me what to do, or I have to do the same refreshing of the mind.(3 votes)

## Video transcript

In the last video, we had a
word problem where we had-- we essentially had to figure out
the sides of a triangle, but instead of, you know, just
being able to do the Pythagorean theorem and because
it was a right triangle, it was just kind of a normal triangle. It wasn't a right triangle. And we just kind of chugged
through it using SOHCAHTOA and just our very simple trig
functions, and we got the right answer. What I want to do now is to
introduce you to something called the law of cosines,
which we essentially proved in the last video, but I want to
kind of prove it in a more-- you know, without the word
problem getting in the way, and I want to show you, once you
know the law of cosines, so you can then apply it to a problem,
like we did in the past, and you'll do it faster. I have a bit of a mixed opinion
about it because I'm not a big fan of memorizing things. You know, when you're 40 years
old, you probably won't have the law of cosines still
memorized, but if you have that ability to start with the trig
functions and just move forward, then you'll
always be set. And I'd be impressed if
you're still doing trig at 40, but who knows? So let's go and let's
see what this law of cosines is all about. So let's say that I
know this angle theta. And let's called this
side-- I don't know, a. No, let's call this side b. I'm being a little
arbitrary here. Actually, let me stay in
the colors of the sides. Let's call that b and let's
call this c, and let's call this side a. So if this is a right triangle,
then we could have used the Pythagorean theorem
somehow, but now we can't. So what do we do? So we know a-- well, let's
assume that we know b, we know c, we know theta, and then
we want to solve for a. But, in general, as long as you
know three of these, you can solve for the fourth once you
know the law of cosines. So how can we do it? Well, we're going to do
it the exact same way we did that last problem. We can drop a line
here to make-- oh, my God, that's messy. I thought I was using
the line tool. Edit, undo. So I can drop a line like that. So I have two right angles. And then once I have right
triangles, then now I can start to use trig functions and
the Pythagorean theorem, et cetera, et cetera. So, let's see, this is a right
angle, this is a right angle. So what is this side here? Let me pick another color. I'm probably going to get too
involved with all the colors, but it's for your improvement. So what is this side here? What is the length of that
side, that purple side? Well, that purple side is just,
you know, we use SOHCAHTOA. I was just going to write
SOHCAHTOA up here. So this purple side is adjacent
to theta, and then this blue or mauve side b is the hypotenuse
of this right triangle. So we know that-- I'm just
going to stick to one color because it'll take me forever
if I keep switching colors. We know that cosine of theta--
let's call this side, let's call this kind of subside--
I don't know, let's call this d, side d. We know that cosine of theta
is equal to d over b, right? And we know b. Or that d is equal to what? It equals b cosine theta. Now, let's call this
side e right here. Well, what's e? Well, e is this whole c
side-- c side, oh, that's interesting-- this whole c side
minus this d side, right? So e is equal to c minus d. We just solved for d, so
side e is equal to c minus b cosine of theta. So that's e. We got e out of the way. Well, what's this magenta
side going to be? Well, let's call this magenta--
let's call it m from magenta. Well, m is opposite to theta. Now, we know it. We've solved for c as well, but
we know b, and b is simple. So what relationship gives us
m over b, or involves the opposite and the hypotenuse? Well, that's sine:
opposite over hypotenuse. So we know that m over b is
equal to sine of theta. We know that-- let
me go over here. m over b, right, because this
is the hypotenuse, is equal to sine of theta, or that m
is equal to b sine of theta, right? So we figured out m, we
figured out e, and now we want to figure out a. And this should
jump out at you. We have two sides of
a right triangle. We want to figure
out the hypotenuse. We can use the
Pythagorean theorem. The Pythagorean theorem tells
us a squared is equal to m squared plus e squared, right? Just the square of
the other two sides. Well, what's m squared
plus e squared? Let me switch to another
color just to be arbitrary. a squared is equal
to m squared. m is b sine of theta. So it's b sine of theta
squared plus e squared. Well, e we figure out is this. So it's plus c minus b
cosine theta squared. Now, let's just chug
through some algebra. So that equals b sine-- b
squared sine squared of theta. Sine squared of theta
just means sine of theta squared, right? Plus, and we just foiled
this out, although I don't like using foil. I just multiply it out. c squared minus 2cb cosine
theta plus b squared cosine theta, right? I just expanded this out
by multiplying it out. And now let's see if we can
do anything interesting. Well, if we take this term and
this term, we get-- those two terms are b squared sine
squared of theta plus b squared cosine-- this should be squared
there, right, because we squared it. b squared cosine squared of
theta, and then we have plus c squared minus 2bc cosine theta. Well, what does
this simplify to? Well, this is the same thing
as b squared times the sine squared theta plus
cosine squared of theta. Something should be jumping out
at you, and that's plus c squared minus 2bc cosine theta. Well, this thing, sine
squared plus cosine squared of any angle is 1. That's one of the
earlier identities. That's the Pythagorean
identity right there. So this equals 1, so then
we're left with-- going back to my original color. We're almost there-- a squared
is equal to-- this term just becomes 1, so b squared. We're just left with a b
squared plus c squared minus 2bc cosine of theta. That's pretty neat, and this
is called the law of cosines. And it's useful because, you
know, if you know an angle and two of the sides of
any triangle, you can now solve for the other side. Or really, if you want to, if
you know three sides of a triangle, you can now solve
for any angle, so that also is very useful. The only reason why I'm a
little bit, you know, here, there, is I don't-- if you are
in trigonometry right now and you might have a test, you
should memorize this because it'll make you faster, and
you'll get the answer right quicker. I'm not a big fan of just
memorizing it without knowing where it came from, because a
year from now or two years from now when you go to college and
it's been four years since you took trigonometry, you probably
won't have this memorized. And if you face a trig problem
all of a sudden, it's good to kind of get there from scratch. With that said, this is the law
of cosines, and if you use the law of cosines, you could have
done that problem we just did a lot faster because we just--
you know, you just have to set up the triangle and then just
substitute into this, and you could have solved for a in
that ship off-course problem. I'll see you in the next video.