# Expanding and Removing Brackets

#### Casio

##### Member
a2 = (c - y)2 + h2

I have been previously advised by another member here that;

(c - y)2 does not equate to c2 - y2

OK with this in mind,

a2 = c2 - 2cy

OK one step at a time.

I squared the C by using 2 outside the bracket, then I moved the 2 inside the bracket and this removed the brackets.

I am assuming that bringing the 2 down I am now subtracting c from C2 thus leaving -2cy.

Is this the case I am not sure?

Then it seems the y2 is then added back.

I originally had (c - y)2

I seem to have ended up with more than I started with?

Kind regards

Casio

#### CaptainBlack

##### Well-known member
a2 = (c - y)2 + h2

I have been previously advised by another member here that;

(c - y)2 does not equate to c2 - y2

OK with this in mind,

a2 = c2 - 2cy

OK one step at a time.

I squared the C by using 2 outside the bracket, then I moved the 2 inside the bracket and this removed the brackets.

I am assuming that bringing the 2 down I am now subtracting c from C2 thus leaving -2cy.

Is this the case I am not sure?

Then it seems the y2 is then added back.

I originally had (c - y)2

I seem to have ended up with more than I started with?

Kind regards

Casio
Please post the question, as it is we will just be guessing at what you are trying to do.

CB

#### Casio

##### Member
Thank you CB for replying. I am looking at modifying Pythagoras Theorem to get to the understanding of cosine rules.

I should have used FOIL, but did not think of it at the time, sorry my fault my health is not brilliant and I am getting old for learning new tricks, I show willing but is difficult from a book sometimes Anyway this is what I have learned.

a2 = (c - y)2 + h2

I want to get rid of the brackets.

a2 = (c - y)(c - y) this equates to;

a2 = c2 - 2cy + y2 +h2

I'll continue on with my learning curve now.

Thanks

#### Sudharaka

##### Well-known member
MHB Math Helper
Re: Algebra a bit of a mind field to me

Thank you CB for replying. I am looking at modifying Pythagoras Theorem to get to the understanding of cosine rules.

I should have used FOIL, but did not think of it at the time, sorry my fault my health is not brilliant and I am getting old for learning new tricks, I show willing but is difficult from a book sometimes Anyway this is what I have learned.

a2 = (c - y)2 + h2

I want to get rid of the brackets.

a2 = (c - y)(c - y) this equates to;

a2 = c2 - 2cy + y2 +h2

I'll continue on with my learning curve now.

Thanks
Hi casio, Firstly I have renamed your thread to make it more relevant to your question. The answer you have obtained after expanding the brackets is correct. Please refer >>this<< for a complete description about expanding brackets.

Kind Regards,
Sudharaka.

#### Casio

##### Member
Thank you Sudharaka for providing the additional information I think I have now managed to understand how to transpose from Pythagoras Theorem to the cosine rule.

I know I have not included a triangle, but image a triangle and put a perpendicular in it to create two right angled triangles.

Take Pythagoras Theorem

b2 = y2 + h2

The letters are changed to reflect the triangle I was using.

previously I got to the stage.

a2 = c2 - 2cy + h2

First I needed to recognise that y + h2 is the same as above in the theorem.

so,

Knowing that Pythagoras Theroem is;

a2 = b2 + c2

I can conclude that;

a2 = b2 + c2 - 2cy

Now comes the really tricky bit, well at least for me anyway

in my triangle ABC, cosA = y / b

That it y = adjacent side and b = hypotenue

so I can say;

b = CosA, and

a2 = b2 + c2 - 2cy becomes

a2 = b2 + c2 - 2bc cosA

So now I understand where the cosine rule has originated from.

Casio 