MHB Expanding and Removing Brackets

[Casio1]

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]

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

 

[Casio1]

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]

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.

 

[Casio1]

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, anda² = b² + c² - 2cy becomesa² = b² + c² - 2bc cosA

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

Casio:D