Proving Trig Identity: csin(A-B/2)=(a-b)cos(C/2)

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Homework Help Overview

The discussion revolves around proving a trigonometric identity related to a triangle ABC, specifically the equation c sin((A-B)/2) = (a-b)cos(C/2). Participants are exploring relationships between the sides and angles of the triangle using trigonometric identities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss a formula from a textbook that resembles the problem and attempt to manipulate it to find a connection to the given identity. There are suggestions to express angle C in terms of angles A and B and to use sine and cosine identities to simplify the expressions. Some participants express uncertainty about their approaches and seek suggestions for further progress.

Discussion Status

The conversation is ongoing, with participants sharing their attempts and questioning the clarity of their expressions. There is no explicit consensus on a method, but some guidance has been offered regarding the use of trigonometric identities and notation.

Contextual Notes

Participants note the importance of proper notation in mathematical expressions, particularly in TeX syntax, and express frustration over clarity in communication. There is a recognition of the challenge in proving the identity, with some participants feeling they are not making progress.

Appleton
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Homework Statement


Given a triangle ABC prove that

c sin \frac{A-B}{2} = (a-b)cos \frac{C}{2}

Homework Equations

The Attempt at a Solution



It looks rather similar to a formula mentioned in my book's lead-in to this exercise:

\frac{a-b}{a+b}=tan\frac{A-B}{2}tan\frac{C}{2}

Which can be rearranged to

(a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}sin\frac{A-B}{2}=(a-b)cos\frac{C}{2}<br />

So from this it would seem I need only prove that

(a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}= c

However this doesn't seem much of a simplification and I feel I'm still at square one trying to prove another identity. I also felt my approach was a bit unorthodox. Any suggestions would be appreciated.
 
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a,b,c are side lengths, A,B,C are the angles?

I didn't find a solution, but expressing C in terms of A and B and converting the cos to a sine makes the two sides more similar. There are also formulas for the sine of a sum/a difference.
 
Thanks for your reply, yes, a,b,c are side lengths and A,B,C are angles.
 
Appleton said:

Homework Statement


Given a triangle ABC prove that

c sin \frac{A-B}{2} = (a-b)cos \frac{C}{2}

Homework Equations

The Attempt at a Solution



It looks rather similar to a formula mentioned in my book's lead-in to this exercise:

\frac{a-b}{a+b}=tan\frac{A-B}{2}tan\frac{C}{2}

Which can be rearranged to

(a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}sin\frac{A-B}{2}=(a-b)cos\frac{C}{2}<br />

So from this it would seem I need only prove that

(a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}= c

However this doesn't seem much of a simplification and I feel I'm still at square one trying to prove another identity. I also felt my approach was a bit unorthodox. Any suggestions would be appreciated.

What do you mean by "csin"?
 
Sorry for any confusion, I ommitted a space. c is the opposite side length to the angle C and is being multiplied by sin ((A-B)/2).
 
Appleton said:
Sorry for any confusion, I ommitted a space. c is the opposite side length to the angle C and is being multiplied by sin ((A-B)/2).
To lessen ambiguity, you can use an asterisk to indicate multiplication -- c * sin((A - B)/2).
 
Please use the respective TeX syntax for trig functions. \sin instead sin, \tan instead of tan etc. trig functions are operators and not variables. It's bad style and uncomfortable to read.
 
nuuskur said:
Please use the respective TeX syntax for trig functions. \sin instead sin, \tan instead of tan etc. trig functions are operators and not variables. It's bad style and uncomfortable to read.
Phooey.
As far as I'm concerned, ##\sin(x)## is only marginally better than ##sin(x)##.

Also, trig functions are not operators - they are functions.
 
Sorry, I meant operators in TeX terms. I did write my work without proper syntax and got booed on by Eeryone. things like sin f(x) , \sin{f(x)}, \frac{d}{dx}f(x) , \frac{\mathrm{d}}{\mathrm{d}x}f(x). Now I boo on it, myself :D Curiously, no one is complaining about the improper use of f(x), but it just hurts my eyes if everything is italic - difficult to keep track of where a function, operator is or a variable.
 
Last edited:
  • #10
nuuskur said:
Sorry, I meant operators in TeX terms. I did write my work without proper syntax and got booed on by Eeryone. things like sin f(x) , \sin{f(x)}, \frac{d}{dx}f(x) , \frac{\mathrm{d}}{\mathrm{d}x}f(x). Now I boo on it, myself :D Curiously, no one is complaining about the improper use of f(x), but it just hurts my eyes if everything is italic - difficult to keep track of where a function, operator is or a variable.
It's not difficult if you use parentheses, as in sin(f(x)). People are going to comment if what you write is unclear.

This one really seems like overkill: \frac{\mathrm{d}}{\mathrm{d}x}f(x). I use as little LaTeX as I can - pages with tons of LaTeX sometimes take much longer to render in a browser, so there's a cost to using it unnecessarily.
 
  • #11
OK I'll bear that in mind next time I post. I'm still no closer to solving it. I tried mfb's method but couldn't progress any further with it. I feel like I'm just trying random tautologies in the hope it leads somewhere, I'm reminded of monkeys and typewriters.
 

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