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Prove trig identity

  • Thread starter Appleton
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  • #1
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Homework Statement


Given a triangle ABC prove that

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

Homework Equations




The Attempt at a Solution



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

[itex]\frac{a-b}{a+b}=tan\frac{A-B}{2}tan\frac{C}{2}[/itex]

Which can be rearranged to

[itex](a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}sin\frac{A-B}{2}=(a-b)cos\frac{C}{2}
[/itex]

So from this it would seem I need only prove that

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

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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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Thanks for your reply, yes, a,b,c are side lengths and A,B,C are angles.
 
  • #4
Ray Vickson
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Homework Statement


Given a triangle ABC prove that

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

Homework Equations




The Attempt at a Solution



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

[itex]\frac{a-b}{a+b}=tan\frac{A-B}{2}tan\frac{C}{2}[/itex]

Which can be rearranged to

[itex](a+b)\frac{sin\frac{C}{2}}{cos\frac{A-B}{2}}sin\frac{A-B}{2}=(a-b)cos\frac{C}{2}
[/itex]

So from this it would seem I need only prove that

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

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"?
 
  • #5
91
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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).
 
  • #6
33,286
4,989
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).
 
  • #7
525
345
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.
 
  • #8
33,286
4,989
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.
 
  • #9
525
345
Sorry, I meant operators in TeX terms. I did write my work without proper syntax and got booed on by Eeryone. things like [itex]sin f(x) , \sin{f(x)}, \frac{d}{dx}f(x) , \frac{\mathrm{d}}{\mathrm{d}x}f(x) [/itex]. Now I boo on it, myself :D Curiously, noone 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
33,286
4,989
Sorry, I meant operators in TeX terms. I did write my work without proper syntax and got booed on by Eeryone. things like [itex]sin f(x) , \sin{f(x)}, \frac{d}{dx}f(x) , \frac{\mathrm{d}}{\mathrm{d}x}f(x) [/itex]. Now I boo on it, myself :D Curiously, noone 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
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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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