Sine rule alternative proof?

Leaving Cert Student

Can I prove the sine rule like this, it seems so much easier than the way in the book:

Given standard triangle with lengths a,b,c and angles across from each side A,B,C.

Use triangle area formula for each angle:
0.5absinc = .5acsinb = .5bcsina ... because all areas are obvioulsy equal.

multiply by two and divide by abc product to get:

sinc/c = sinb/b = sina/a

too easy surely?

magicianz

Whats your logical reasoning behind what you're doing, other than "it works" ?

Plucking numbers out of the air will mean you won't get the marks in Leaving Cert.

Leaving Cert Student

magicianz wrote: »

Whats your logical reasoning behind what you're doing, other than "it works" ?

Plucking numbers out of the air will mean you won't get the marks in Leaving Cert.

Haha wait are kidding? I hardly plucked them from mid air. There is clear logical progression, I am simply wondering is it a recognized proof? Thanks though that made my day hahaha.

Leaving Cert Student

In case you werent joking, I used the triangle area formula for each angle, equated them because the area isn't changing. Then I simply tidied up to get it how I wanted.

Prodigious

Looks good to me, but then again I wasn't even aware we had to prove the sine rule :P

Leaving Cert Student

Methememb wrote: »

Looks good to me, but then again I wasn't even aware we had to prove the sine rule :P

To be fair there isn't a proper list of the proofs we need to know anywhere and people seem to have different answers... :-\

MathsTeacherD

It's a decent proof (as in, it's perfectly general and not a special case or anything) but where it falls short is that it relies on the formula for the area of a triangle, and you haven't proven that that is true. So, you haven't plucked the number 0.5 out of the air, etc, but you've plucked the formula...

Trigonometry proofs are usually done without relying on any but the most basic of assumptions, and everything is proven before your very eyes. For the LC-HL syllabus, you need to be able to derive trig formulae 1-7, 9 from the 24 on the syllabus. You can use formulae 1-3 to help you prove 4, etc. The sine rule is number 2, so you've basically got to be able to do it from scratch!

Leaving Cert Student

MathsTeacherD wrote: »

It's a decent proof (as in, it's perfectly general and not a special case or anything) but where it falls short is that it relies on the formula for the area of a triangle, and you haven't proven that that is true. So, you haven't plucked the number 0.5 out of the air, etc, but you've plucked the formula...

Trigonometry proofs are usually done without relying on any but the most basic of assumptions, and everything is proven before your very eyes. For the LC-HL syllabus, you need to be able to derive trig formulae 1-7, 9 from the 24 on the syllabus. You can use formulae 1-3 to help you prove 4, etc. The sine rule is number 2, so you've basically got to be able to do it from scratch!

Ok so just to confirm? This proof would be unacceptable?

MathsTeacherD

Well, I wouldn't use it...

Leaving Cert Student

MathsTeacherD wrote: »

Well, I wouldn't use it...

Ok cheers... frustrating that there isnt a clear list of them somewhere, and this proof I found on a leaving cert tutorials channel on youtube!

user.name

Does anyone know of a list of proofs that we need to know? They are trying to get away from the whole learning and reproducing proofs in project maths but they still ask them ugh :pac:

MathsManiac

You could make the proof complete by first of all proving that area formula, which only takes a few lines.

Leaving Cert Student

MathsManiac wrote: »

You could make the proof complete by first of all proving that area formula, which only takes a few lines.

True, where could I find that proof? is it in our textbooks?

Leaving Cert Student

ray giraffe wrote: »

[latex]\sin{A}=\frac{h}{a} \sin{B} =\frac{h}{b}[/latex]
[latex]a \sin{A} = b \sin {B}[/latex]
[latex]\frac{a}{\sin{A}}=\frac{b}{\sin{B}}[/latex]

Lovely, I assume this proves it for all triangles too... Now why couldn't they have put that in the textbook, so simpe.

Thank you very much sir!

ray giraffe

3 line proof of Sine Rule (right this time!)



[latex]\displaystyle \sin {A}=\frac{h}{b}, \sin{B} =\frac{h}{a}[/latex]

[latex]\displaystyle b \sin{A} = h = a \sin {B}[/latex]

[latex]\displaystyle \frac{a}{\sin{A}} = \frac{b}{\sin{B}}[/latex]

Leaving Cert Student

ray giraffe wrote: »

Sorry just realised I messed it up - new version coming!

Yep, a and b mixed up... its grand though I get it now, taking it down in maths copy...

Leaving Cert Student

Cheers!