# Trig challenge problem: arithmetic progression

## Video transcript

if the angles A, B, and C of a triangle
are in an arithmetic progression and if a, b and c - lower case a, b and c
denote the length of the sides opposite to the capital, the angles A - capital A, capital
B and capital C respectively then what is the value of this expression
right over here so let's see if we can work our way through this, so let's just draw the triangle so
we have visualization of what all the letters represent so we have the angles A, B and C so let me just draw them like this so we have the angles A B and C and then the sides opposites them are the
lower case versions, so the side opposite the capital A is lower case a the side opposite to capital B is lower case
b and the side opposite to the angle capital C is lower case c. Now the first piece of information
they tell us is that they the angles capital A, capital B and capital C of the
triangle are in an arithmetic progression arithmetic progression very fancy word but all an arithmetic progression is, is a series of numbers
that are separated by the same amount, so and let me give you some examples, so 1,2,3 - that's an arithmetic progression 2, 4, 6 arithmetic progression were separated by 2
every time i could do 10, 20, 30 also an arithmetic progression. These are all
arithmetic progressions. So all they're saying is is to go from angle A to angle B - however much
that is, is this the same amount to go from angle B to angle C. So let's see what that tells us what that tells us about or maybe tells us, maybe doesn't tell us anything about those angles so if we could say we could say we have angle A and then we have the notion angle B so we could say that B is equal to A plus some constant we don't know what that is. it could go by 1, it could go by 2, it could go by 10 we don't know what it is, so A plus N and then C would be equal to B plus N which is the same thing which is the same thing as B is A plus N, so this is A plus N
it plus plus N which is equal to A plus 2N so what does that tell us? Well, the other thing
we know about three angles in a triangle that they have to add up to a 180
degrees. So: this, this and this have to add up to a 180 degrees let's try it out so we have A plus A plus N plus A plus 2N, plus A plus 2N is going to be equal to 180 degrees we have one, two, three "A"s here, so we get 3A plus with one N and another two N: three A plus 3 N is equal to 180 degrees or you could divide both sides by 3 and you
get A plus N is equal to A plus N is equal to 60 degrees so what does that what does that tell us well A could still be anything cause if N is 1 then A is 59 if N is 10 then A is going to be 50, so it doesn't give us much information about the angle A but if you look up here, do you see A+N anywhere? you see it right over here. B is equal to
A plus N and we just figured out that A plus N has to be equal to sixty degrees so using this first piece of information we
are able to come up with something pretty tangible B must be equal to sixty degrees and you
could try it out with a bunch of numbers these could be 59 60 and 61 that's an arithmetic progression and once again B is the middle one
right over here. These could be 50 60 and 70 could be 40 60 and 80 but no matter what the arithmetic progression
is in order for these three angles to add up to 180 the middle one has to be equal to has to be equal to 60 degrees. So that was
a pretty that's.. we're doing pretty well so far so let's
see what we can do at the next part with the next part of problem. I'm trying to save some screen real estate right over here okay so they want us to figure out the value of
the expression a over c sine of two C - capital C, plus c over a sine of two A. So
let me just write it down so we have i'll do it.. i'll do it in blue a over c a over c sine of two times capital C plus c over a sine of two times capital A what's that going to be equal to? so, whenever
you see stuff like this you got a 2 here a 2 here frankly the best things you should do is just experiment
with your trigonometric identities and see if anything pops out of you that might be
useful and a little bit of a clue here the first
part of the problem helped us figure out what B is it helped us figure out what B is but right now the expression has no B in it.
So right now this information seems kind of useless but if we could put this somehow in terms of B then we'll have will will be making progress
as we know information about angle B so let's see what we can do so the first thing i would use is well sine of 2A - let me just rewrite each
of these so sine of i just say sine of two times anything that's just the same thing as and this is called the double angle formula so this is i might be wrong i was for the actual
names of them but sine of two time something is two sine of that something times the cosine of that times the cosine of that something and you'll
see that in any trigonometric book on the inside cover even a lot of calculus books let's do that for this the same thing right
over here so sine of 2A over here is going to be 2 sine of A cosine of A that's just a standard trigonometric identity and we, in the trigonometric, less we prove
that identity i think we do it multiple times then out on front we have our coefficients till we have a over c times this plus c over a times this now is there anything we can do and remember
in the back of our mind we should be thinking of how can we use this information that be is
equal to sixty so if we can somehow put this in the form got a b here when i think about how do you get a b here
I think well you know we have a triangle here so the things
that relate the sides of the triangle when especially when it's not a right triangle we are really gonna deal with the law of sines or law of cosines and a law of sines let me just rewrite it over
here just for our reference Law of sines would say sine of A over a is equal to sine of B over b which is equal to sine of C over c And it looks like we might be able to use that
let me just read the law of cosine here the law of cosine is c squared it should look as a Pythagorean theorem with little
adjustment for the fact that is not a right triangle So c squared is equal to a squared plus b
squared minus 2 ab cosine cosine of C of capital C so it's law of sines and law of cosines, so let's if we can somehow use both of these to put these in
terms of b which we have information about well the first thing is i could rewrite this so this is sine of C
over c and this is sine of A over a, so let me do that so i have so i have the 2a I have 2a cosine of c let me write that separately so that's 2a cosine cosine of capital c and then times sine of C over c times with white sine of c that's a capital c sine of capital c over lower case c, that's
that term and that term right over there and then to that I'm adding to that I'm adding let me do this the same thing over
here, I have two times I'm going to separate these guys out Actually I wanted to do the sines so let me separate I'm gonna separate this guy and in this guy out and so I get plus 2c cosine of A times sine of capital a over lower case a times sine of capital a over lower case a and what does this
do for me Well look at the law of sines
right over there I have sine of C over c that's that over there and then I've sine of A over a and that's that over
there capital a over lower case a. they're both equal to sign of B over b so we're making progress we started introducing b into the equation and that's what we actually have information about so this It could be rewritten a sine of be over b so this is the same thing as sine of capital b over lower case b and this is the same thing a sine of capital b over lower case b and they're both being multiplied or both of these terms are multiplying are being multiplied by that 2a cosine of capital c times that and then plus 2c it's a lower case c cosine of capital a times that so we can factor out the sine of B over b so let's do that, let's factor it out so this is the same thing as it's the same thing as 2a 2a and I really have a sense of what the next
step is so I'll leave a space here 2a times cosine of C plus this and these are being multiplied I'll have
some space there plus 2c two lower case c times the cosine of A and all of this all of this times the sine the sine of B over b and we already know that B is 60 degrees so we can evaluate this uh... pat pretty easily but let's just continue let's see if we can somehow put this right over here in terms of b well if you look over here we have 2a cosine of C 2c cosine of A it's trying to look pretty darn close each of these terms look preety darn close to this part to this part of the law of cosines over there and actually let's solve for that part of the law of cosines let's see what we could do so if you add if you add 2ab cosine C to both sides you get 2ab cosine of capital c plus c squared is equal to a squared plus b squared or if you subtract c squared from both sides you get 2ab cosine of capital c is equal to a squared plus b squared
minus c squared and this is interesting and you know we can switch around the letters later on but this looks pretty darn close to the so what if and this looks pretty darn close to this except here we're dealing with
an a instead of the c, we've just switched letters around and we could rewrite this actually let me rewrite it just for fun i could rewrite this over here as 2 2cb not rewrite I can swap the letters times the cosine of A you see I'm swaping the a(s) and the c(s) is equal to c squared plus b squared minus a squared there's nothing unique about the sides I can do this with all of the sides so here It's a big c here, you have an a and b out front and then you have tha a squared plus b squared minus the small c squared, if you have the big A then you gonna have the cb in the front and you're subtracting the a squared right over here and this is useful because this term right over here this term right
over here looks almost like this term over here if we could just
multiply this by b so let's do that we can multiply that by b but let's multiply this whole numerator this whole term by b(s)
if we multiply this whole term by b what do we get we get a b there we get a b there and of course you can just arbitrarily multiply an expression by b that'll change its value. so what we could do is multiply the expression by b which we just did we distributed the b across here but they were also divided by b where which so I'll divide by b, that's an equivalent
of multiplying the denominator the denominator there not b squared that's equivalent of multiplying the denominator there by b. That simple thing is dividing it by b We've multiplied it by b divided by b That's the same thing I'm just turning this into b squared now what does this give us well we have this term right over here this term right over here is now the exact
same thing is that over there so it is now in a squared plus b squared
minus c squared and then this term right over here is now the exact same thing as this thing over here which is the same thing as that We're using the law of cosines so this is plus c squared plus b squared minus a squared and then all of that times this sine of b sine of capital b over b squared now what does this give us the we have an a squared and a negative a squared things are starting to simplify a squared negative a squared we have a negative c squared and a positive c squared so what are we left with we're just left out with the 2b squared so our whole expression has simplified to 2b squared sine of B sine of capital b over lower case b squared these cancel out so our whole expression simplifies to 2 sine of B and from from the get go we knew what b was we know it's sixty degrees so this is equal to two times the sine of sixty degrees and if you
don't have the sine of sixty degrees memorized you can always just break out of thirty sixty
ninety triangle so let me draw this is a right triangle right over here this is
sixty degrees hypotenuse has length one we're dealing with
the unit circle this side is thirty degrees the side opposite to the thirty degrees is one-half the side opposite to the sixty degrees is square root of three times that so it's a sqaure root of three over two You can even use the Pythagorean theorem to figure out when you know one of them you could figure out the other one So it's the sine of, sinus opposite over hypotenuse so square root of three
over two over one Or just square root of three over two, so this is equal to two times there's a home stretch, it;s very exciting ! square root of three over two these cancel out so we are left with the square root of three that's a pretty neat problem and just in case you're curious this came from the two thousand ten IIT joint entrance examination hard to get into engineering and
science universities in India and they gave this example like you know hundreds of thousands of kids and you know the top the top i don't know like two thousand actually get into one of the ideas but anyway I just thought it was a pretty pretty neat problem