a maths question (trig) (1 Viewer)

[littleboy]

can you smarrt people help me with this question?

a) By repeated application of factoring by the difference of squares, prove the identity

(2ab)^2 - (a^2 + b^2 - c^2)^2 = (a+b+c)(a+b-c)(a-b+c)(-a+b+c)


B) Let triangle ABC be any triangle, and let s=1/2(a+b+c) be the semiperimeter. Prove that:


(a+b+c)(a+b-c)(a-b+c)(-a+b+c) = 16s(s-a)(s-B)(s-c)


c)Write down the formula for cosC in terms of the sides a,b and c, then use question a) and B) and the Pythagorean identities to prove that sinC =

{2 x root[s(s-a)(s-B)(s-c]}/ab

d) hence show that the area of the triangle is

root[s(s-a)(s-B)(s-c)]

thanks

 

[Kulazzi]

honey, I suggest you post in the maths forum. You'd get a lot of help there because there would also be yr 12, uni students viewing that forum

 

[Trev]

(2ab)² - (a² + b² - c²)²
= [2ab - (a² + b² - c²)][2ab + (a² + b² - c²)]
= [2ab - a² + b² - c²][2ab + a² + b² - c²]
.... [2ab - a² + b² - c²] equals (a-b+c)(-a+b+c)
and.....[2ab + a² + b² - c²] equals (a+b+c)(a+b-c)
therefore...
[2ab - a² + b² - c²][2ab + a² + b² - c²] = (a+b+c)(a+b-c)(a-b+c)(-a+b+c)

 

[Kulazzi]

well done Trev, now you can just ignore my advice......

 

[Trev]

haha soz Kulazzi =P i feel stoopd i duno wat to do for the otha q haha *runs in shame*

 

[Kulazzi]

thats ok, at least you solved something

 

[Trev]

heres no.2....
(a+b+c)(a+b-c)(a-b+c)(-a+b+c) = 16s(s-a)(s-B)(s-c)
since s=1/2(a+b+c) then 2s = (a+b+c)
therefore..
a+b+c=2s
a+b-c=2s-2c
a-b+c=2s-2b
-a+b+c=2s-2a
......(a+b+c)(a+b-c)(a-b+c)(-a+b+c) = 2s(2s-2c)(2s-2b)(2s-2a)
= 2.2.2.2(s-c)(s-b)(s-a)
=16(s-c)(s-b)(s-a)
yay!

 

[Trev]

rep me for doing these darnt!

 

[Trev]

heres d), havnt done c) yet....
since
sinC = {2 x root[s(s-a)(s-B)(s-c]}/ab
ab.sinC = 2.root[s(s-a)(s-B)(s-c)]
(absinC)/2 = area formula of abc triangle, therefore
(absinC)/2 = root[s(s-a)(s-B)(s-c)]

 

[Trev]

cosC = (a² + b² - c²)/2ab
cos²C + sin²C = 1
therefore sin²C = 1 - cos²C
from before, cos²C = [(a² + b² - c²)/2ab]²
thefore sin²C = 1 - [(a² + b² - c²)/2ab]²
....
addition leaves
sin²C = [(2ab)² - (a² + b² - c²)²]/(2ab)²
since (2ab)² - (a² + b² - c²)² (from part a.) = (a+b+c)(a+b-c)(a-b+c)(-a+b+c)
and since (a+b+c)(a+b-c)(a-b+c)(-a+b+c) (from part b.) = 16s(s-a)(s-B)(s-c)
....
sin²C = [16s(s-a)(s-B)(s-c)]/(2ab)²
sin²C = [4.root{s(s-a)(s-B)(s-c)}]/2ab
sinC = 2.root{s(s-a)(s-B)(s-c)}]/ab
note: sinC is only positive because area of triangle cannot be negative (ie. root of x² = +x and -x; in this case the answer is only positive...
REP ME!