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Expanding and Removing Brackets

Casio

Member
Feb 11, 2012
86
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?

Could somebody please explain it?

Kind regards

Casio
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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?

Could somebody please explain it?

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
Feb 11, 2012
86
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(Wondering)

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
Feb 5, 2012
1,621
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(Wondering)

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
Feb 11, 2012
86
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:D